Proof $f(x) = e^{-3x} + 7\cos(6x)-50x+34$ bijective and $(f^{-1})'(42)$

Question

Let $f: \mathbb{R} \to \mathbb{R}$ with

$$f(x) = e^{-3x} + 7\cos(6x)-50x+34$$

It says that this function is bijective and I can see that with WolframAlpha but how can one prove that?

And when I tried calculating $(f^{-1})'(42)$ no online math tool can solve it. Why?

Answer 1

Just note that $f'(x)=-3^{-3x}- 42 \sin(6x) -50$ is always negative (easy if you look at each term). A strictly decreasing function is injective. If you also note that $f$ is continuous and $$ \lim_{x \to -\infty} f(x)= +\infty, \quad \lim_{x \to +\infty} f(x)= -\infty $$ you see that $f$ is surjective. At this point you know that $f$ is bijective. Finally,

$$ (f^{-1})'(42) = \frac{1}{f'(c)}, $$

where $c$ is such that $f(c)=42$, i.e. $c=0$. This means that $$ (f^{-1})'(42) = \frac{1}{f'(0)}=-\frac{1}{53} $$

Answer 2

$$f'(x)=-3e^{-3x}-42(\sin x+1) -8 <0$$ so $f(x)$ is monotonically decreasinf for all real numbers so it is one to one and $f(-\infty)>0$ and $f(\infty) <0$.\infty} so $f: R\rightarrow R$ is on to.We know that $f(0)=42$ so $f^{-1}(42)=0$ Then $$\frac{df^{-1}(y)}{dy}|_{y=42}=\frac{1}{f'(0)}=-\frac{1}{53}.$$