2 logarithmic questions: Two inequalities and one inverse related

Question

I have attached the image below depicting the problems and my solutions or what I had attempted so far.

For Q3(a):3cosx = 2x for -π < x < π (less than or equal to sign) [I attempted formatting as Tex command - hope this works] $$ 3\cos{x} = 2x, \text{for} -\pi \leq x \leq \pi $$

\begin{align*} \text{Graphically the point of intersection is :} 0.915 \end{align*}

(However when I check by inputting the value of 0.915 in the original functions, both answers are different - for example: e.g. $3\cos{0.915} = 2.999$ vs. $2(0.915) = 1.83$) \begin{align*} \ 3cos(0.915) \ = 2.999 \end{align*} \begin{align*} \ vs. \end{align*} \begin{align*} \ 2(0.915) \ = \ 1.83 \end{align*}

Q8: I think I did the inverse part correctly and composed $f^{-1}(x)$ into $f(x)$ correctly. I am not sure how to proceed to verify that the inverse is equivalent to $x$.

Once again, thank you very much for help. Best community right here!

Answer 1

For the second question $f(x)=3\log(x+1)$

The inverse of $f(x)=f^{-1}(x)=e^{\frac x3}-1$

Now $f(f^{-1}(x))=3\log(e^{\frac x3}-1+1)=x$

Answer 2

For the first question, the solution is effectively $x\approx 0.915$ but the result is in radians.

Checking $$3 \cos(0.915)-2\times 0.915=-0.000628274$$ while, if, by mistake, your calculator is set for degrees, $3 \cos (0.915 {}^{\circ})=2.99962$.