Question #0321b

Question

1177 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-06T08:22:18

See explanation.

a. Calculate the inverse function.

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

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

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

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

b. Calculate the derrivative

$g'(x)=1/(3*root(3)((x+18)^2))$

c. Substitute $x=9$

$g'(9)=1/(3root(3)((9+18)^2))=1/(3root(3)((27)^2))$

$g'(9)=1/(3root(3)(729))=1/(3*9)=1/27$

Answer 2 · 2018-02-06T11:29:53

$1/27$ .

Since, $g(x)=f^-1(x), g(f(x))=x$ .

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

$g'(f(x))*f'(x)=1................(star)$ .

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

$:. x=3$ .

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

$g'(f(3))*f'(3)=1...............(starstar)$ .

But,

$f(x)=x^3-18 rArr f'(x)=3x^2 rArr f'(3)=27, and f(3)=9$ .

Utilising these in $(starstar)$ , we finally have,

$g'(9)=1/(f'(3))=1/27$ .

Enjoy Maths.!