Answers to Calculus Practical 6

Question 1 §

Finding the critical points of the following functions…

Question 1a §

$$f(x)=x^{\frac{7}{8}}(x-3{)}^2:x\ge 0$$

$\text{Differentiating this function using the product and chain rules,}$

$$f^{\prime }(x)=\frac{7}{8}{x}^{-\frac{1}{8}}(x-3{)}^2+2x^{\frac{7}{8}}(x-3):x\ge 0$$

$$⟹f^{\prime }(x)=(x-3)\left[\frac{7}{8}{x}^{-\frac{1}{8}}(x-3)+2x^{\frac{7}{8}}\right]$$

$\text{Solving for when }{f}^{\prime }(x)=0\text{ to find the critical points (as the gradient would be 0),}$

$$(x-3)\left[\frac{7}{8}{x}^{-\frac{1}{8}}(x-3)+2x^{\frac{7}{8}}\right]=0$$

$$⟹x=0,\frac{21}{23},3$$

$\text{After substituting these x values into the original equation, we find}$

$$\text{Critical points }(x,y)=(0,0),\left(\frac{21}{23},\frac{230421^{\frac{7}{8}}}{52923^{\frac{7}{8}}}\right),(3,0)$$

Question 1b §

$$f(x)=\frac{x-1}{x^2+4}$$

$$\text{Differentiating this function by using the quotient rule}$$

$$f^{\prime }(x)=\frac{2x(x-1)-(x^2+4)}{(x^2+4{)}^2}$$

$$⟹f^{\prime }(x)=\frac{x^2-2x-4}{(x^2+4{)}^2}$$

$\text{Solving for when }{f}^{\prime }(x)=0\text{ to find the critical points (as the gradient would be 0),}$

$$f^{\prime }=0:x^2-2x-4=0$$

$$⟹x=1\pm \sqrt{5}\text{ by using the quadratic formula}$$

$\text{After substituting these x values into the original equation, we find}$

$$\text{Critical points }(x,y)=\left(1+\sqrt{5},\frac{\sqrt{5}-1}{8}\right),\left(1-\sqrt{5},-\frac{\sqrt{5}+1}{8}\right)$$

Question 2 §

Finding the absolute maxima and minima of the following functions…

Question 2a §

$$f(x)=2x^3+3x^2-12x-12\in [-3,2]$$

$$⟹f^{\prime }(x)=6x^2+6x-12=0\text{gradient}=0\text{ at extrema}$$

$$⟹x^2+x-2=0:x=-2,1$$

$$f^{\prime \prime }(x)=12x+6:f^{\prime \prime }(-2)<0,f^{\prime \prime }(1)>0$$

$$\therefore \begin{array}{c}\text{Absolute maxima at }(-2,8)\\ \text{Absolute minima at }(1,-19)\end{array}\text{for the given interval }[-3,2]$$

Question 2b §

$$f(x)=x+\frac{1}{x}\in [0.2,4]$$

$$⟹f^{\prime }(x)=1-\frac{1}{x^2}=0\text{gradient}=0\text{ at extrema}$$

$$⟹x^2-1=0:x=\pm 1:x=-1\text{ is not in the interval}$$

$$f^{\prime \prime }(x)=\frac{2}{x^3}:f^{\prime \prime }(1)>0$$

$$\therefore \text{Absolute minima at }(1,2)\text{ for the given interval }[0.2,4]$$

Question 3 §

Find the intervals of increase and decrease of the following functions…

Question 3a §

$$f(x)=x^3-3x+2$$

$$⟹f^{\prime }(x)=3x^2-3\text{gradient}=0\text{ at extrema}$$

$$⟹x^2-1=0:x=\pm 1$$

$$\therefore \text{Three intervals of increase/decrease: }(-\infty ,-1),(-1,1),(1,\infty )$$

$$\text{Testing one value from each interval,}$$

$$\begin{array}{c}(-\infty ,-1)⟹f^{\prime }(-2)=9>0⟹\text{increasing}\\ (-1,1)⟹f^{\prime }(0)=-3<0⟹\text{decreasing}\\ (1,\infty )⟹f^{\prime }(2)=9>0⟹\text{increasing}\end{array}$$

Question 3b §

$$g(x)=x\cdot e^{-\frac{3x^2}{2}}$$

$$⟹g^{\prime }(x)=e^{-\frac{3x^2}{2}}-\frac{3x^2}{3}\cdot e^{-\frac{3x^2}{2}}\text{gradient }=0\text{ at extrema}$$

$$⟹e^{-\frac{3x^2}{2}}\left(1-x^2\right)=0:x=\pm 1$$

$$\therefore \text{Three intervals of increase/decrease: }(-\infty ,-1),(-1,1),(1,\infty )$$

$$\text{Testing one value from each interval,}$$

$$\begin{array}{c}(-\infty ,-1)⟹f^{\prime }(-2)=-3e^{-6}<0⟹\text{decreasing}\\ (-1,1)⟹f^{\prime }(0)=1>0⟹\text{increasing}\\ (1,\infty )⟹f^{\prime }(2)=-3e^{-6}<0⟹\text{decreasing}\end{array}$$

Question 4 §

To find a formula for ${\tanh }^{-1}(x)$,

$$\tanh (x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}=y$$

$\text{Then by solving for x,}$

$$e^x-e^{-x}=y(e^x+e^{-x})$$

$$⟹e^x-e^{-x}=y{e}^x+y{e}^{-x}$$

$$⟹(1-y)e^x-(1+y)e^{-x}=0$$

$$⟹(1-y)e^{2x}-(1+y)=0$$

$$⟹e^{2x}=\frac{1+y}{1-y}$$

$$⟹x=\frac{1}{2}\ln \left(\frac{1+y}{1-y}\right)$$

$$\therefore {\tanh }^{-1}(x)=\frac{1}{2}\ln \left(\frac{1+x}{1-x}\right):\begin{array}{c}\text{domain: }\\ \text{range: }\end{array}$$