Question

(a) Find the derivative of the function 𝑦 = 2𝑥^2+12/x^2, when 𝑥 = 2.

(b) Let 𝑓(𝑥) = −3/𝑥−7. Find the inverse of the function.

Answer 1

Solution (a)

$𝑦 = 2𝑥^2+\frac{12}{x^2}$

$𝑦 = 2𝑥^2+12x^{-2}$

$\frac{dy}{dx}= (2)(2)𝑥^{2-1}+(12)(-2)x^{{-2-1}}$

$\frac{dy}{dx}= 4𝑥-24x^{-3}$

at the point $x=2$

$\frac{dy}{dx}= 4(2)-(24)(2)^{-3}$

$\frac{dy}{dx}= 12-\frac{24}{8}$

$\frac{dy}{dx}= 9$

Hence the derivative of $𝑦 = 2𝑥^2+12/x^2$, when $𝑥 = 2$ is

$\frac{dy}{dx}= 9$

Solution (b)

Given that $𝑓(𝑥) =y= -\frac{3}{x-7}$ .

Interchanging $x$ (domain) and $y$ (range)

$x= -\frac{3}{y-7}$

Now making $y$ the subject of the formula, we have

$x(y-7)=-3$

$y-7=-\frac{3}{x}$

$y=-\frac{3}{x}+7$

Hence the inverse function of the given function $𝑓(𝑥) =y= -\frac{3}{x-7}$ is

$𝑓^{-1}(𝑥) =-\frac{3}{x}+7$