Another funny function

Using the chain rule, we have that

$g'(x) = \dfrac{1}{f(e^{x})} * f'(e^{x}) * \dfrac{d}{dx}(e^{x})$ .

Thus $g'(0) = \dfrac{1}{f(1)} * f'(1) * e^{0} = \dfrac{1}{2} * 3 * 1 = \dfrac{3}{2} = \boxed{1.5}$ .

The first step is solving for f(x).

We know the slope of f is equal to 3.

f(x)=3x+b

We also know that the f(1)=2

2=3(1)+b=2-3=b

b=-1

f(x)=3x-1

We then find the f(e^x) by substituting e^x for x.

f(e^x)=3e^x-1

We then plug this equation in for the f(e^x).

g(x)=ln(3e^x-1)

We then must use chain rule to get the derivative.

The derivative for ln(u) where u is another equation is (1/u)(u')

g'(x)=(3e^x)/(3e^x-1)

g'(0)=(3e^0)/(3e^0-1)=3/(3-1)=3/2=1.5