Question

Hello

Can you help me to do these two exercises please I block on those please it is on the derivatives of a number

Thank you so much in advance!!!

Answer 1

First rememeber that if we have a function like:

g(x) = x^n

then:

dg(x)/dx = g'(x) = n*x^(n - 1)

And also if we have:

g(x) = f(x) + h(x)

then:

dg(x)/dx = g'(x) = f'(x) + h'(x)

1) We have the function f(x) = (x + 1)^2

We can rewrite this as:

f(x) = x^2 + 2*x + 1

Using both things written above, we know that:

f'(x) = 2*x + 1*2 = 2*x + 2

And we want to find f'(-2), so we only need to evaluate the above function in x = -2, this is:

f'(-2) = 2*-2 + 2 = -4 + 2 = -2

f'(-2) = -2

b) I suppose that this refers to the inverse function of f(x)

An inverse function is such that:

g( f(x)) = x

f( g(x)) = x

For f(x) = (x + 1)^2

The inverse will be something that first cancels that square, and then subtracts 1.

Then:

g(x) = √x - 1

if we evaluate this in f(x) we get:

g( f(x)) = √(f(x)) - 1 = √(x + 1)^2 - 1 = (x + 1) - 1 = x

Then the inverse function of f(x) is:

g(x) = √x - 1

This can also be written as:

g(x) = x^(1/2) - 1

Then the derivative of this will be:

g'(x) = (1/2)*x^(1/2 - 1) = (1/2)*x^(-1/2) = (1/2)*(1/√x)