# Question #0321b

Feb 6, 2018

See explanation.

#### Explanation:

a. Calculate the inverse function.

$y = {x}^{3} - 18$

$y + 18 = {x}^{3}$

$x = \sqrt[3]{y + 18}$

So the inverse function is: $g \left(x\right) = \sqrt[3]{x + 18}$

b. Calculate the derrivative

$g ' \left(x\right) = \frac{1}{3 \cdot \sqrt[3]{{\left(x + 18\right)}^{2}}}$

c. Substitute $x = 9$

$g ' \left(9\right) = \frac{1}{3 \sqrt[3]{{\left(9 + 18\right)}^{2}}} = \frac{1}{3 \sqrt[3]{{\left(27\right)}^{2}}}$

$g ' \left(9\right) = \frac{1}{3 \sqrt[3]{729}} = \frac{1}{3 \cdot 9} = \frac{1}{27}$

Feb 6, 2018

$\frac{1}{27}$.

#### Explanation:

Since, $g \left(x\right) = {f}^{-} 1 \left(x\right) , g \left(f \left(x\right)\right) = x$.

Differentiating w.r.t. $x$, using the Chain Rule, we have,

$g ' \left(f \left(x\right)\right) \cdot f ' \left(x\right) = 1. \ldots \ldots \ldots \ldots \ldots \left(\star\right)$.

For $g ' \left(9\right)$, we must have, here, $f \left(x\right) = 9 , i . e . , {x}^{3} - 18 = 9$.

$\therefore x = 3$.

Sub.ing $x = 3$ in $\left(\star\right)$, we have,

$g ' \left(f \left(3\right)\right) \cdot f ' \left(3\right) = 1. \ldots \ldots \ldots \ldots . . \left(\star \star\right)$.

But,

$f \left(x\right) = {x}^{3} - 18 \Rightarrow f ' \left(x\right) = 3 {x}^{2} \Rightarrow f ' \left(3\right) = 27 , \mathmr{and} f \left(3\right) = 9$.

Utilising these in $\left(\star \star\right)$, we finally have,

$g ' \left(9\right) = \frac{1}{f ' \left(3\right)} = \frac{1}{27}$.

Enjoy Maths.!