Inverse Function Theorem (Domain, one-to-one, application)

Question

$f(x) = e^x + \ln(x)$

Q1) Find the Domain

Q2) Show that $f$ is one-to-one

Q3) Find $(f^{-1})(e)$

Q4) Find the derivative of the inverse function at $e$.

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I am pretty much at a loss for how to apply the IFT.

Q1)

I know the domain of $f^{-1}$ = range of $f$ 

I know that the range of $f^{-1}$ = domain of $f$

Q2)

I know that I need to check the signs of the derivative and if it is 
always positive then it must be one-to-one but why is that the case?

Q3 & Q4)

I think this is where the theorem is supposed to be applied just not 
sure how.

If someone could walk me through how to solve these problems utilizing the IFT that would be appreciated very much. Thanks!

Answer 1

Q_1): Since $\ln$ is only defined for $x>0$, $f$ has the domain $(0, \infty)$.

Q_2): For $x>0$ we have $f'(x)=e^x+ \frac{1}{x} >0$. Conclusion ?

Q_3): $f(1)=e$, conclusion ?

Q_4): $(f^{-1})'(e)=\frac{1}{f'(1)}= ?$

Answer 2

The first part is asking for the domain of $f(x)=e^x+\ln(x)$. Recall that $e^x$ is defined for all $x\in\Bbb R$, but $\ln(x)$ is only defined for $x>0$. The domain of the function is the intersection of the domains of $e^x$ and $\ln(x)$, which is $x>0$ or $\Bbb R^+$.

As for the second part, as you have correctly pointed out, you need to differentiate $f$ and see the sign of the derivative. If the derivative of a differentiable function is always positive or always negative, the function is injective.

To see why, recall that Rolle's Theorem tells us for a function $h(x)$ continuous in $[a,b]$ and differentiable in $(a,b), h(a)=h(b)\implies\exists c\in(a,b)$ such that $f'(c)=0$. Take the contrapositive of this: $\forall c\in(a,b),f'(c)\ne0\implies h(a)\ne h(b)$. If the derivative is either always positive or always negative in $(a,b)$, it is never zero and $h(a)\ne h(b)$. You can extend this to any arbitray $a,b$ in the domain.

For the third part, you can easily inspect $f(1)=e$.

For the last part, recall that $f(f^{-1}(x))=x\implies f'(f^{-1}(x))\cdot (f^{-1})'(x)=1$. Putting $x=e$, we get,

$$f'(1)\cdot(f^{-1})'(e)=1$$

Can you solve it further?