End of chapter exercises

Textbook Exercise 6.13

Determine ${f}'\left(x\right)$ from first principles if $f\left(x\right)=2x-{x}^{2}$.

\begin{align*} f(x) & = 2x - x^{2}\\ f(x+h) & = -(x+h)^{2}+2(x+h)\\ f'(x) &= \lim_{h \to 0}\dfrac{f(x+h)-f(x)}{h} \\ &=\lim_{h \to 0}\dfrac{-(x+h)^{2} + 2(x+h) -(-x^{2} + 2x)}{h}\\ &=\lim_{h \to 0}\dfrac{-x^{2} - 2xh - h^{2} + 2x + 2h + x^{2} - 2x}{h}\\ &=\lim_{h \to 0}\dfrac{-2xh - h^{2} + 2h}{h}\\ &=\lim_{h \to 0}\dfrac{h(-2x - h + 2)}{h}\\ &=\lim_{h \to 0}(-2x - h + 2)\\ &= -2x + 2 \end{align*}

Given $f\left(x\right)=\frac{1}{x}+3$, find ${f}'\left(x\right)$ using the definition of the derivative.

\begin{align*} f'(x) &=\lim_{h \to 0} \frac{\left(\frac{1}{x+h}+3\right)-\left(\frac{1}{x}+3\right)}{h}\\ &=\lim_{h \to 0} \frac{\frac{1+3(x+h)}{x+h}-\frac{1+3x}{x}}{h}\\ &=\lim_{h \to 0} \frac{\frac{x(1+3x+3h) - (x+h)(1+3x)}{x(x+h)}}{h}\\ &=\lim_{h \to 0} \frac{x(1+3x+3h)-(x+h)(1+3x)}{h \cdot x(x+h)}\\ &=\lim_{h \to 0} \frac{x +3x^{2}+3xh-x - 3x^{2} - h - 3xh}{hx(x+h)}\\ &=\lim_{h \to 0} \frac{-h}{hx(x+h)}\\ &=\lim_{h \to 0} \frac{-1}{x^{2}+xh}\\ &= -\frac{1}{x^{2}} \end{align*}

Calculate: $\displaystyle\lim_{x\to 1}\frac{1-{x}^{3}}{1-x}$

\begin{align*} \lim_{x \to 1} \frac{x^{3}-1}{x-1} & = \lim_{x \to 1} \frac{(x-1)(x^{2}+x+1)}{x-1} \\ & = \lim_{x \to 1} (x^{2} + x + 1)\\ &= 3 \end{align*}

$y=(x+2)(7-5x)$

\begin{align*} y &=14-3x-5x^{2} \\ \frac{dy}{dx} &= -3-10x \end{align*}

$y=\frac{8x^{3}+1}{2x+1}$

\begin{align*} y & = \frac{8x^{3}+1}{2x+1} \\ & = \frac{(2x+1)(4x^{2}-2x+1)}{2x+1} \\ & = 4x^{2}-2x+1 \\ \frac{dy}{dx} &= 8x-2 \end{align*}

$y={\left(2x\right)}^{2}-\frac{1}{3x}$

\begin{align*} f(x) & = (2x)^{2} - \frac{1}{3x}\\ f(x) & = 4x^{2} - \frac{x^{-1}}{3} \\ f'(x) & = 8x + \frac{x^{-2}}{3} \\ f'(x) & = 8x + \frac{1}{3x^{2}} \end{align*}

$y=\frac{2\sqrt{x}-5}{\sqrt{x}}$

\begin{align*} f(x) & = \frac{2\sqrt{x}-5}{\sqrt{x}}\\ & = \frac{2x^{\frac{1}{2}}}{x^{\frac{1}{2}}} - \frac{5}{x^{\frac{1}{2}}}\\ & = 2 - 5x^{\frac{-1}{2}}\\ f'(x) & = -5(\frac{-1}{2})x^{\frac{-3}{2}}\\ & = \frac{5}{2\sqrt{x^{3}}} \end{align*}

Use the definition of the derivative to calculate ${f}'\left(x\right)$.

\begin{align*} f(x) & = 2x^{2} - x \\ f(x + h) & = 2(x + h)^{2} - (x + h) \\ & = 2(x^{2} + 2xh + h^{2}) - x - h \\ & = 2x^{2} + 4xh + 2h^{2} - x - h \\ & \\ f'(x) &=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ &=\lim_{h \to 0} \frac{(2x^{2} + 4xh + 2h^{2} - x - h)-( 2x^{2} - x)}{h}\\ &=\lim_{h \to 0} \frac{2x^{2} + 4xh + 2h^{2} - x - h - 2x^{2} + x}{h}\\ &=\lim_{h \to 0} \frac{4xh + 2h^{2} - h }{h}\\ &=\lim_{h \to 0} \frac{h(4x + 2h - 1) }{h}\\ &=\lim_{h \to 0} (4x + 2h - 1)\\ &= 4x - 1 \end{align*}

Hence, calculate the coordinates of the point at which the gradient of the tangent to the graph of $f$ is $\text{7}$.

\begin{align*} m & = 4x - 1 = 7 \\ 0 & = 4x - 8 \\ 0 & = x - 2 \\ \therefore x &= 2 \end{align*}

The $y$ value is: \begin{align*} f(x) & = 2x^{2} - x \\ \therefore f(2) & = 2(2)^{2} - 2\\ & = 8 - 2\\ & = 6 \end{align*} The coordinates of the point are $(\text{2};\text{6}).$

If $g\left(x\right)={\left({x}^{-2}+{x}^{2}\right)}^{2}$, calculate ${g}'\left(2\right)$.

\begin{align*} g(x) & = \left(x^{-2}+x^{2}\right)\left(x^{-2}+x^{2}\right) \\ & = x^{-4} + x^{0} +x^{0} +x^{4}\\ & = x^{-4} + 1 + 1 +x^{4}\\ & = x^{-4} + 2 + x^{4}\\ g'(x) & = -4x^{-5}+4x^{3}\\ g'(2) & = -4(2)^{-5}+4(2)^{3}\\ &=\frac{-4}{32} + 32\\ &=-\frac{1}{8}+32 \\ &=31\frac{7}{8} \\ &=\frac{255}{8} \end{align*}

Find ${f}^{-1}\left(x\right)$.

\begin{align*} y & = 2x-3\\ f^{-1}: \enspace x & = 2y-3 \\ 2y& = x+3 \\ y&=\frac{x}{2}+\frac{3}{2} \end{align*}

Solve ${f}^{-1}\left(x\right)=3{f}'\left(x\right)$.

\begin{align*} \frac{x+3}{2} & = 3(2)\\ x+3&=12\\ & = 9 \end{align*}

$p(t)=\frac{\sqrt[5]{{t}^{3}}}{3}+10$

\begin{align*} p(t) & = \frac{\sqrt[5]{t^{3}}}{3} + 10\\ p(t) &= \frac{t^{\frac{3}{5}}}{3} + 10\\ p'(t) & = \frac{1}{3}\left(\frac{3}{5}t^{-\frac{2}{5}}\right)\\ &=\frac{1}{5\sqrt[5]{t^{2}}} \end{align*}

$k(n)=\frac{\left(2{n}^{2}-5\right)\left(3n+2\right)}{{n}^{2}}$

\begin{align*} k(n) &=\frac{6n^{3}+4n^{2}-15n-10}{n^{2}}\\ &=6n+4-15n^{-1}-10n^{-2}\\ k'(n) &= 6 + 15n^{-2} + 20n^{-3}\\ &= 6 + \frac{15}{n^{2}} + \frac{20}{n^{3}} \end{align*}

If $xy-5=\sqrt{{x}^{3}}$, determine $\frac{dy}{dx}$.

\begin{align*} xy - 5 & = \sqrt{x^{3}}\\ y & = \frac{x^{\frac{3}{2}}+5}{x}\\ & = x^{\frac{1}{2}}+5x^{-1}\\ \frac{dy}{dx} & = \frac{1}{2}x^{-\frac{1}{2}} - 5x^{-2}\\ &=\frac{1}{2\sqrt{x}}-\frac{5}{x^{2}} \end{align*}

Determine $\frac{dy}{dx}$.

Differentiate $y$ with respect to $x$.

\[\frac{dy}{dx}=3x^{2}\]

Find $\frac{dx}{dy}$.

To differentiate $x$ with respect to $y$, express $x$ in terms of $y$:

\begin{align*} x&=\sqrt[3]{y}\\ &=y^{\frac{1}{3}}\\ \frac{dx}{dy}&=\frac{1}{3}y^{-\frac{2}{3}}\\ &=\frac{1}{3\sqrt[3]{y^{2}}} \end{align*}

Show that $\frac{dy}{dx} \times \frac{dx}{dy} = 1$.

\begin{align*} \frac{dy}{dx} \times \frac{dx}{dy} &= 3x^{2} \times \frac{y^{-\frac{2}{3}}}{3} \\ \text{But } y&=x^{3} \\ \therefore \frac{dy}{dx} \times \frac{dx}{dy} &= 3x^{2} \times \frac{\left(x^{3}\right)^{-\frac{2}{3}}}{3} \\ &=x^{2} \times x^{-2} \\ &=x^{0} \\ &=1 \end{align*}

Calculate $f\left(-1\right)$.

\begin{align*} f(-1) & = (-1)^{3} -3(-1)^{2} + 4\\ & = \text{0} \end{align*}

Hence, solve $f\left(x\right)=0$.

We know that $(x + 1)$ is a factor of $f(x)$ because $f(-1) = 0$. We factorise further using inspection:

\begin{align*} f(x) & = (x+1)(x^{2} - 4x + 4) \\ & = (x+1)(x-2)(x-2) \\ \therefore \enspace f(x)&=(x+1)(x-2)(x-2) \\ 0 &=(x+1)(x-2)(x-2) \\ \therefore \enspace x = -1 & \text{ or } x=2 \end{align*}

Determine ${f}'\left(x\right)$.

\begin{align*} f'(x) & = 3x^{2} - 6x\\ & = 3x(x-2) \end{align*}

Sketch the graph of $f$, showing the coordinates of the turning points and the intercepts on both axes.

The $y$-intercept is $y=4$.

The $x$-intercepts are $(2;0) \text{ and } (-1;0)$.

To find the turning points we let the derivative equal $0$. $f'(x)=3x(x-2)=0$. The $x$-values of the turning points are: $x=0$ and $x=2$.

Therefore, the turning points are $(0;4)$ and $(2;0)$.

Determine the coordinates of the points on the graph of $f$ where the gradient is $\text{9}$.

\begin{align*} f'(x)& = 3x^{2} - 6x \\ 3x^{2} - 6x &= 9\\ 3x^{2} - 6x - 9& = 0 \\ x^{2} - 2x -3 &=0 \\ (x-3)(x+1)&=0\\ \therefore \enspace x=3 \text{ or } x&=-1 \\ \text{If } x = 3: \enspace f(3) & ={(3)}^{3}-3{(3)}^{2}+4 \\ & = 27 -27 +4 \\ & = 4 \\ \text{If } x = -1: \enspace f(-1) & ={(-1)}^{3}-3{(-1)}^{2}+4 \\ & = -1 - 3 + 4 \\ &= 0 \end{align*}

Therefore, at $(-1;0)$ and $(3;4)$.

Draw the graph of $f'(x)$ on the same system of axes.

\begin{align*} f'(x) &=3x^{2}-6x \\ y_{\text{int}}:& \enspace(0;0) \\ x_{\text{int}}:& \enspace 3x(x-2)=0 \\ \therefore \enspace x&=0 \text{ or } x=2 \\ \therefore (0;0) & \text{ or } (2;0) \\ & \\ \text{Turning point }&\text{ is where } f''(x)=0 \\ \therefore \enspace&6x-6=0 \\ \therefore \enspace x&=1\\ \text{Turning point: }&(1;-3) \end{align*}

Determine $f''(x)$ and use this to make conclusions about the concavity of $f$.

\begin{align*} f''(x) &=6x-6 \\ 6x-6&=0 \text{ for } x=1 \\ x<1, \enspace f''(x)<0&\therefore f(x) \text{ is concave down} \\ x>1, \enspace f''(x)>0&\therefore f(x) \text{ is concave up} \\ x=1, \enspace f''(x)=0&\therefore \enspace f(x) \text{ is point of inflection} \end{align*}

If $f(-1)=0$, determine the $x$-intercepts of $f$.

\begin{align*} f\left(x\right)&=2{x}^{3}-5{x}^{2}-4x+3 \\ f(-1)&=0 \text{, so } (x+1) \text{ is a factor of } f(x) \\ f(x) &= (x+1)(2x^{2}-7x+3) \\ &=(x+1)(2x-1)(x-3) \\ \text{Let } \enspace f(x)&=0 \\ \therefore \enspace&(x+1)(2x-1)(x-3)=0\\ \therefore \enspace x&=-1, \enspace x=\frac{1}{2} \text{ or } x=3 \end{align*}

The $x$-intercepts of $f$ are $\left(-1;0\right)$, $\left(\frac{1}{2};0\right)$ and $\left(3;0\right)$.

Determine the coordinates of the turning points of $f$.

\begin{align*} f'(x) & = 6x^{2} -10x - 4\\ 0 & = 2(3x^{2} - 5x - 2)\\ & = 3x^{2} - 5x - 2 = 0\\ &= (3x+1)(x-2)=0\\ \therefore \enspace x&=-\frac{1}{3} \text{ or } x=2 \\ \text{Substitute } x&= -\frac{1}{3}:\\ y&=2\left(-\frac{1}{3}\right)^{3}-5\left(-\frac{1}{3}\right)^{2}-4\left(-\frac{1}{3}\right)+3 \\ &=\frac{100}{27} \\ &=\text{3,7} \\ \text{Substitute } x&= 2:\\ y&=2(2)^{3} -5(2)^{2}-4(2)+3 \\ &=-9 \\ \therefore \text{ Turning points are }& A\left(-\frac{1}{3}; \text{3,7}\right) \text{ and } B\left(2;-9\right) \end{align*}

Draw a sketch graph of $f$. Clearly indicate the coordinates of the turning points and the intercepts with the axes.

For which value(s) of $k$ will the equation $f\left(x\right)=k$ have three real roots of which two are equal?

To find the point when the cubic function has two real roots, we need to find the points where one of the turning points lies on the $x$-axis. We look at the $y$-values of the turning points to determine this. For the maximal turning point we lower the graph and for the minimal one we raise the graph. This gives: $k=\text{3,7}$ or $k=-\text{9}$

Determine the equation of the tangent to the graph of $f\left(x\right)=2{x}^{3}-5{x}^{2}-4x+3$ at the point where $x=1$.

Find the $y$-value at $x=1$: \begin{align*} f(1) & = 2(1)^{3}-5(1)^{2}-4(1)+ 3\\ & = -4 \end{align*} Determine the value of $m$ substituting $x=1$ into the derivative: \begin{align*} f'(x) & = 6x^{2} - 10x - 4 \\ f'(1) & = 6(1)^{2} - 10(1) - 4 \\ m & = -8 \end{align*}

The equation of the tangent is: \begin{align*} y - y_{1} & = m(x - x_{1}) \\ y + 4 & = -8(x - 1) \\ y & = -8x + 4 \end{align*}

Show by calculation that $b=9$, $c=27$ and $d=26$.

\begin{align*} y_{\text{int}} (0;26): \enspace & d=26 \\ \therefore f(x) &= x^{3} + bx^{2} +cx + 26 \\ x_{\text{int}} (-2;0): \enspace f(-2)&=(-2)^{3} + b(-2)^{2} + c(-2) +26 \\ \therefore 0 &=-8+4b-2c+26 \\ 0 &=4b-2c+18 \\ 2b-c&=-9 \quad \textbf{ eqn 1}\\ \text{Stationary point at } x&=-3 \\ \therefore f'(-3) &=0 \\ f'(x) &=3x^{2}+2bx+c \\ f'(-3) &= 3(-3)^{2} +2b(-3) +c \\ 0 &=27 - 6b +c \\ 6b-c&=27 \quad \textbf{ eqn 2}\\ 2b-c&=-9 \quad \textbf{ eqn 1}\\ \textbf{ eqn 1 } - \textbf{ eqn 2 }: \enspace 4b &=36 \\ b&=9\\ \text{Substitute } b&=9 \text{ into } \enspace \textbf{ eqn 1} \\ 2(9)-c&=-9\\ -c&-27 \\ c&=27\\ \therefore f(x) &=x^{3} + 9x^{2} +27x + 26 \end{align*}

Find the $y$-coordinate of the point of inflection.

\begin{align*} \text{Passes through }(-3;y) \\ \therefore y &= (-3)^{3} + 9(-3)^{2} + 27(-3) +26 \\ &= -27 + 81 -81 + 26 \\ &=-1 \\ \therefore \text{ Point of inflection: }&A(-3;-1) \end{align*}

Draw the graph of $f$.

Discuss the gradient of $f$.

Gradient always positive except at $A(-3;-1)$ where the gradient is at its minimum value, zero.

Discuss the concavity of $f$.

Concavity changes at inflection point $A(-3;-1)$:

\begin{align*} f'(x)&=3x^{2}+18x+27 \\ f''(x):&=6x+18 \\ x<-3, \enspace f''(x)<0&\therefore f(x) \text{ is concave down} \\ x=-3, \enspace f''(x)=0&\therefore \enspace f(x) \text{ is point of inflection} \\ x>1, \enspace f''(x)>0&\therefore f(x) \text{ is concave up} \end{align*}

Identify the stationary points of the cubic function, $g(x)$.

Turning points are at $x = -1$ and $x = 3$:

For $x = -1$: this is a local maximum turning point (the gradient of $g$ changes from positive to negative).

For $x = 3$: this is a local minimum turning point (the gradient of $g$ changes from negative to positive).

What is the gradient of function $g$

where $x=0$.

At $x = 0$, the gradient of the function is equal to $-\text{2}$. We can write this as $g'(0)= -2$.

If it is further given that $g(x)$ has only two real roots, draw a rough sketch of $g(x)$. Intercept values do not need to be shown.

Given that $h(x)$ is a linear function with $h(2)=11$ and $h'(2)=-1$, find the equation of $h(x)$.

$h(x)$ is a linear function and is therefore of the form $y = mx + c$. The derivative of a linear function is $\frac{dy}{dx} = m$, therefore from $h'(2) = -1$, we know that $m = -1$.

\begin{align*} \therefore y &= -x + c \\ \text{Substitute } (2;11): \enspace 11 &= -2 + c \\ \therefore c &= 13 \\ \therefore y &= -x + 13 \end{align*}

Therefore, $h(x) = -x + 13$.

Use the graphs and determine the values of $x$ for which:

$f(x)$ is a decreasing function.
$f(x) \cdot g(x) \geq 0$.
$f'(x)$ and $g(x)$ are both negative.

$x < -1$ and $x > 3$
$x = -3$ or $x \geq 3$ (both $f$ and $g$ are negative)
$f'(x) < 0$ where the gradient of $f$ is negative. Therefore, $x < -1$ or $x > 3$.

And $g < 0$ for $x < -3$ or $x > 3$.

Therefore, $f'$ and $g < 0$ for $x < -3$ or $x > 3$.

Given $f(x)=-x^{3}+3x^{2}+9x-27$, determine the equation of the tangent to $f$ at the point $E (2;y)$.

Equation of tangent to $f$ at $E(2;y)$:

\begin{align*} f'(x) &= -3x^{2} + 6x + 9 \\ \text{At } x=2: \enspace f'(2) &= -3(2)^{2} + 6(2) + 9 \\ &= -12 +12 + 9 \\ &= 9 \\ \therefore y &= 9x + c \end{align*}

To calculate the value of $y$ at $E$, substitute $x = 2$ into $f(x)$:

\begin{align*} f(2) & =-(2)^{3}+3(2)^{2}+9(2)-27 \\ & =-8 + 12 + 18 -27 \\ & =-5 \end{align*}

Substitute $(2;-5)$ into $f'(x)$ to determine $c$:

\begin{align*} y &= 9x + c\\ -5 & = 9(2) + c \\ \therefore c & = -23 \\ y &= 9x - 23 \end{align*}

Find the coordinates of the point(s) where the tangent in the question above meets the graph of $f$ again.

Equate the equation of the tangent and $f(x)$:

\begin{align*} 9x - 23 &= -x^{3}+3x^{2}+9x-27 \\ \therefore 0 & = -x^{3}+3x^{2} - 4 \\ \text{Let } k(x) & = -x^{3}+3x^{2} - 4 \\ k(-1) & = -(-1)^{3}+3(-1)^{2} - 4 \\ &= 1 + 3 -4 \\ &= 0 \\ \therefore k(x) & = (x + 1)(- x^2 + 4x - 4) \\ & = - (x + 1)(x^2 - 4x + 4) \\ & = - (x + 1)(x - 2)^{2} \\ \therefore 0 & = - (x + 1)(x - 2)^{2} \\ \therefore x = -1 & \text{ or } x = 2 \text{ or } x = 2 \end{align*}

Therefore, tangent meets graph of $f$ at the turning point $C(-1;-32)$.

Without any calculations, give the $x$-intercepts of the graph of $f'(x)$. Explain reasoning.

$x$-intercepts of $f'$ are at $(-1;0)$ and $(3;0)$, the turning points of $f$ (points $B$ and $C$). At these two points the gradient of the graph is equal to zero, where $f'(x)$ cuts the $x$-axis.

Sketch the graph of $f\left(x\right)={x}^{3}-9{x}^{2}+24x-20$, show all intercepts with the axes and turning points.

We find the $y$-intercept by finding the value for $f(0)$.

\begin{align*} f(x) & = x^{3} -9x^{2} + 24x - 20 \\ f(0) & = (0)^{3} - 9(0)^{2} + 24(0) - 20\\ & = -20 \end{align*}

The $y$-intercept is: $(0;-20)$

We find the $x$-intercepts by finding the values for which the function $f(x) = 0$.

We use the factor theorem to check if $(x-1)$ is a factor.

\begin{align*} f(x) & = x^{3} -9x^{2} + 24x - 20 \\ f(1) & = (1)^{3} - 9(1)^{2} + 24(1) - 20 \\ & = -4 \end{align*}

Therefore, $(x-1)$ is not a factor.

We now use the factor theorem to check if $(x+1)$ is a factor.

\begin{align*} f(x) & = x^{3} -9x^{2} + 24x - 20 \\ f(-1) & = (-1)^{3} - 9(-1)^{2} + 24(-1) - 20 \\ & = -54 \end{align*}

Therefore, $(x+1)$ is not a factor.

We now try $(x-2)$:

\begin{align*} f(x) & = x^{3} -9x^{2} + 24x - 20 \\ f(1) & = (2)^{3} - 9(2)^{2} + 24(2) - 20 \\ & = 0 \end{align*}

Therefore, $(x-2)$ is a factor.

If we divide $f(x)$ by $(x-2)$ we are left with: \[f(x) = (x-2)(x^{2} - 7x + 10)\]

This has factors: \[f(x) = (x-2)(x-5)(x-2)\]

The $x$-intercepts are: $(2;0), (5;0)$.

Find the turning points by setting $f'(x)=0$.

If we use the rules of differentiation we get:

\begin{align*} f'(x) & = 3x^{2} - 18x + 24 \\ 0 & = 3(x^{2} - 6x + 8) \\ & = 3(x-2)(x-4) \end{align*}

The $x$-coordinates of the turning points are: $x=4$ and $x=2$.

The $y$-coordinates of the turning points are calculated as: \begin{align*} f(2) & = (2)^{3} - 9(2)^{2} + 24(2) - 20 \\ & = 0 \end{align*} and \begin{align*} f(4) & = (4)^{3} - 9(4)^{2} + 24(4) - 20 \\ & = -4 \end{align*}

Therefore the turning points are: $(2;0)$ and $(4;-4)$.

Find the equation of the tangent to $f\left(x\right)$ at $x=4$.

We know that at $x=4$, $y=-4$ (this is a turning point of the graph).

We substitute $x=4$ into the derivative of the function to get $m$: \begin{align*} m & = 3(4-2)(4-4) \\ m & = 0 \end{align*}

Substituting this and the coordinates of the point into: $y-y_{1} = m(x-x_{1})$ gives: \begin{align*} y - (-4) & = 0(x-4)\\ y & = -4 \end{align*}

The equation of the tangent to the graph at $x=4$ is $y=-4$.

Determine the point of inflection and discuss the concavity of $f$.

Point of inflection:

\begin{align*} f'(x) & = 3x^{2} - 18x + 24 \\ f''(x) & = 6x - 18 \\ \therefore 0 &= 6x - 18 \\ 6x &= 18 \\ \therefore x &= 3 \\ \text{Substitute } x = 3: \enspace f(3) & = (3)^{3} -9(3)^{2} + 24(3) - 20 \\ & = 27 - 81 + 72 - 20 \\ & = -2 \end{align*}

This gives the point $(3;-2)$.

Concavity:

$f''(x) < 0$ for $x < 3$: concave down

$f''(x) = 0$ for $x = 3$: point of inflection

$f''(x) > 0$ for $x > 3$: concave up

Determine the minimum value of the sum of a positive number and its reciprocal.

Let the number be $x$ and the reciprocal be $\frac{1}{x}$. The sum of these two numbers is $S = x + \frac{1}{x}$. To find the minimum value we must differentiate the expression for the sum and set it equal to zero: \begin{align*} S &=x + \frac{1}{x}\\ S' &= 1 - \frac{1}{x^{2}}\\ 0 &= 1 - \frac{1}{x^{2}}\\ 1 &= \frac{1}{x^{2}}\\ x^{2} & = 1\\ x & = 1 \\ \therefore S &=1 + \frac{1}{1} \\ &= 2 \end{align*} Therefore, the minimum value for the sum of a positive number and it's reciprocal is $\text{2}$.

Calculate the height of the water level in the kettle just before it starts to boil.

The kettle starts to boil at $t = \text{0}$: \begin{align*} d &= 86 -\frac{1}{8} t - \frac{1}{4}t^{3} \\ &= 86 - \frac{1}{8}(0) - \frac{1}{4}(0)^{3}\\ &= \text{86}\text{ mm} \end{align*}

As the water boils, the water level in the kettle decreases. Determine the rate at which the water level is decreasing when $t = \text{2}\text{ minutes}$.

To find the rate at which the water level is decreasing we take the derivative of the depth: \begin{align*} d(t) &= 86 -\frac{1}{8} t - \frac{1}{4}t^{3} \\ d'(t) &= - \frac{1}{8} - \frac{1}{4} (3)t^{2}\\ &= - \frac{1}{8} - \frac{3}{4} t^{2}\\ d'(2) &= - \frac{1}{8} - \frac{3}{4}(2)^{2}\\ &= - \frac{1}{8} - 3\\ &= - \frac{25}{8} \\ &=-\text{3,125}\text{ mm per minute} \end{align*} This rate is negative since the water level is decreasing.

How many minutes after the kettle starts to boil will the water level be decreasing at a rate of $12\frac{1}{8}$ $\text{mm}$ per minute?

To find the number of minutes when the rate will be $-\text{12,125}$ $\text{mm.min$^{-1}$}$ (water level is decreasing) we set the derivative equal to this and solve for $t$: \begin{align*} d'(t) &= - \frac{1}{8} - \frac{3}{4}t^{2}\\ -\text{12,125} &= - \frac{1}{8} - \frac{3}{4}t^{2}\\ \frac{97}{8} - \frac{1}{8}&= \frac{3}{4}t^{2}\\ \frac{3}{4}t^{2}&= 12\\ t^{2} &= 16\\ t &= 4 \end{align*} After $\text{4}$ minutes the water level will be decreasing at $\text{12,125}$ $\text{mm.min$^{-1}$}$.

The displacement of a moving object is represented by the equation: \begin{align*} D(t)&=\frac{4}{3}t^{3}-3t \\ \text{where } D&= \text{ distance travelled in metres} \\ t &= \text{ time in seconds} \end{align*}

Calculate the acceleration of the object after $\text{3}$ seconds.

\begin{align*} D(t)&=\frac{4}{3}t^{3}-3t \\ D'(t)&= 4t^{2} - 3 \\ D''(t)&= 8t \\ \therefore D''(3)&= 8(3) \\ &= 24 \end{align*}

After $\text{3}$ seconds, the acceleration is $\text{24}$ $\text{m.s$^{-2}$}$.

In the figure $PQ$ is the diameter of the semi-circle $PRQ$. The sum of the lengths of $PR$ and $QR$ is $\text{10}$ $\text{units}$. Calculate the perimeter of $\triangle PQR$ when $\triangle PQR$ covers the maximum area in the semi-circle. Leave the answer in simplified surd form.

Let $PR$ be $x$ units and $QR$ be $(10 - x)$ units.

\begin{align*} \text{Area } \triangle PQR &= \frac{1}{2} PR \cdot QR \quad (\angle \text{ in semi-circle})\\ \therefore A(x) &= \frac{1}{2} x(10 - x) \\ &= 5x - \frac{1}{2}x^2 \end{align*}

To find maximum area, let $A'(x) = 0$:

\begin{align*} A'(x) &= 5 - x \\ 0 &= 5 - x \\ \therefore x &= 5 \end{align*}

Therefore, $PR = 5$ units and $QR = 5$ units.

\begin{align*} PQ^{2} &= 5^2 + 5^2 \quad (\text{Pythagoras}) \\ &= 50 \\ \therefore PQ &= \sqrt{50} \\ &= 5\sqrt{2} \end{align*} \begin{align*} \text{Perimeter } \triangle PQR &= 5 + 5 + 5\sqrt{2} \\ &= 5 + 5 + 5\sqrt{2} \\ &= 10+ 5\sqrt{2} \text{ units} \end{align*}

Perimeter of the triangle is $10+ 5\sqrt{2}$ units.

The capacity of a cylindrical water tank is $\text{1 000}$ $\text{litres}$. Let the height be $H$ and the radius be $r$. The material used for the bottom of the tank is twice as thick and also twice as expensive as the material used for the curved part of the tank and the top of the tank.

Remember: $\text{1 000}\text{ ℓ}=\text{1}\text{ m}^{3}$

Express $H$ in terms of $r$.

\begin{align*} V &= \pi r^2 H \\ 1 &= \pi r^2 H \\ \therefore H &= \frac{1}{\pi r^2} \end{align*}

Show that the cost of the material for the tank can be expressed as: \[C=3\pi r^{2}+\frac{2}{r}\]

\begin{align*} \text{Cost of material} &= \left( 2 \times \text{area bottom} \right) + \text{area top} + \text{curved part} \\ &= \left( 2 \times \pi r^{2} \right) + \pi r^{2} + (2\pi r H) \\ &= 3 \pi r^{2} + 2\pi r H \\ \text{Substitute } H &= \frac{1}{\pi r^2}: \\ \text{Cost of material} &= 3 \pi r^{2} + \left( 2\pi r \times \frac{1}{\pi r^2} \right) \\ &= 3 \pi r^{2} + \frac{2}{r} \end{align*}

Determine the diameter of the tank that gives the minimum cost of the materials.

[IEB, 2006]

Let the cost of the material be $C(r)$.

\begin{align*} C(r) &= 3 \pi r^{2} + \frac{2}{r} \\ C'(r) &= 6 \pi r - \frac{2}{r^2} \\ \therefore 0 &= 6 \pi r - \frac{2}{r^2} \\ r^{3} &= \frac{2}{6 \pi} \\ \therefore r &= \text{0,47}\text{ m} \\ \therefore d &= 2 \times \text{0,47}\text{ m} \\ &= \text{0,94}\text{ m} \end{align*}

Determine the volume of the cone in terms of $h$ and $d$.

(Volume of a cone: $V=\frac{1}{3}\pi r^{3}h$)

\begin{align*} r &= \frac{d}{2} \\ h + d &= 10\\ \therefore h &= 10 - d \\ V &= \frac{1}{3} \pi r^2 h \\ &= \frac{1}{3} \pi \left( \frac{d}{2} \right)^2 (10 - d) \\ &= \frac{1}{12} \pi d^{2} (10 - d) \\ &= \frac{1}{12} (10 \pi d^{2} - \pi d^{3}) \end{align*}

Determine the radius and height of the cone for the volume to be a maximum.

Let the volume of the cone be $V(d)$.

\begin{align*} V(d) &= \frac{1}{12} (10 \pi d^{2} - \pi d^{3}) \\ V'(d) &= \frac{1}{12} (20 \pi d - 3 \pi d^{2}) \\ \therefore 0 &= \frac{1}{12} (20 \pi d - 3 \pi d^{2}) \\ 0 &= 20 \pi d - 3 \pi d^{2} \\ 0 &= \pi d( 20 - 3 d) \\ \therefore d = 0 & \text{ or } d = \frac{20}{3} \end{align*} \begin{align*} \therefore d &= \text{6,67}\text{ cm} \\ \therefore r &= \frac{1}{2} \times \frac{20}{3} \\ &= \frac{10}{3} \\ &= \text{3,34}\text{ cm} \\ h &= 10 - \frac{20}{3} \\ &= \frac{10}{3} \\ &= \text{3,34}\text{ cm} \end{align*}

Calculate the maximum volume of the cone.

\begin{align*} V &= \frac{1}{3} \pi r^2 h \\ &= \frac{1}{3} \pi \left( \frac{10}{3} \right) ^2 \left( \frac{10}{3} \right) \\ &= \frac{1}{3} \pi \left( \frac{100}{9} \right) \left( \frac{10}{3} \right) \\ &= \frac{1000 \pi }{81} \\ &= \text{38,79}\text{ cm$^{3}$} \end{align*}

A water reservoir has both an inlet and an outlet pipe to regulate the depth of the water in the reservoir. The depth is given by the function: \begin{align*} D(h)&= 3 +\frac{1}{2}h - \frac{1}{4}h^{3} \\ \text{where } D&= \text{ depth in metres} \\ h&= \text{ hours after 06h00} \end{align*}

Determine the rate at which the depth of the water is changing at 10h00.

\begin{align*} D(h)&= 3 +\frac{1}{2}h - \frac{1}{4}h^{3} \\ D'(h)&= \frac{1}{2} - \frac{3}{4}h^{2} \\ \therefore D'(4)&= \frac{1}{2} - \frac{3}{4}(4)^{2} \\ &= \frac{1}{2} - 12 \\ &= -11 \frac{1}{2} \end{align*}

Therefore, rate of change of the depth of the water is $-\text{11,5}$ $\text{m}$ per hour at 10h00.

Is the depth of the water increasing or decreasing?

Decreasing (rate is negative).

At what time will the inflow of water be the same as the outflow?

[IEB, 2006]

\begin{align*} D'(h)&= \frac{1}{2} - \frac{3}{4}h^{2} \\ 0 &= \frac{1}{2} - \frac{3}{4}h^{2} \\ \frac{3}{4}h^{2} &= \frac{1}{2} \\ h^{2} &= \frac{2}{3} \\ \therefore h &= \text{0,82} \text{ hours} \end{align*}

We can convert this to minutes: $\text{0,82} \times \text{60} \approx \text{49}$ minutes. Therefore, at 06h49

6.8 Summary

Table of Contents

7.1 Revision