Question #0321b

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Answer 1 · 2018-02-06T08:22:18

See explanation.

a. Calculate the inverse function.

#y=x^3-18#

#y+18=x^3#

#x=root(3)(y+18)#

So the inverse function is: #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.!