Here is a look at what I am doing. The I can evaluate the first one no problem. The second one is causing me problems and the solution given isn't helping me to get my head around it. Can someone please try and give me the steps and an explanation? Thanks
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For the second problem, you are given \begin{align} f_{1}(x) &=3x+2 \\ f_{2}(x) &=4x^3 \end{align} and you are asked to find $$(f_{1}^{-1}\circ f_{2})(1.9).$$ To do this, you will need to first find the inverse function of $f_{1}(x)$. To do that, we can take the equation $y = 3x+2$ and solve for $y$: \begin{align} y &= 3x + 2\\ y - 2 &=3x\\ \frac{1}{3}(y - 2) &= x. \end{align} This tells you that the inverse function is given by the equation $$f_{1}^{-1}(y) = \frac{1}{3}(y-2),$$ or to keep the input notation the same with all your functions, we can write this as $$f_{1}^{-1}(x) = \frac{1}{3}(x-2)$$ as they did in the solution. Then, to evaluate \begin{align} (f_{1}^{-1}\circ f_{2})(1.9) &= f_{1}^{-1}(f_{2}(1.9)). \end{align} This means we first need to evaluate $f_{2}(1.9)$. This gives us $$f_{2}(1.9) = 4(1.9)^{3} = 27.436,$$ which leaves us with $f_{1}^{-1}(27.436)$. To evaluate that, we use the inverse function we found, so $$f_{1}^{-1}(27.436) = \frac{1}{3}(27.436 - 2) = 8.478$$ to three decimal places. So, all in all, we have that $$(f_{1}^{-1}\circ f_{2})(1.9) = 8.478$$ to three decimal places.
If $y=f_1(x)=3x+2$, then $y-2=3x$ and $x=\frac13(y-2)$, so $f^{-1}(y)=\frac13(y-2).$
To calculate $f_1^{-1}(f_2(1.9))$, first calculate $f_2(1.9)=4\times1.9^3=27.436$.
Now calculate $f_1^{-1}(f_2(1.9))=f_1^{-1}(27.436)=\frac13(27.436-2)\approx8.479.$