Here is a section of the question:
It is given that there is exactly one real value of t satisfying the equation $t = \sqrt[3]{(t + 0.8)}$. Verify by calculation that this value lies between 1.2 and 1.3.
I do know how to answer this but i'm just curious why do you have to let $$f(t) = t - \sqrt[3]{(t+0.8)}$$ before substituting the values in? Why not just let $$f(t) = \sqrt[3] {(t+0.8)}?$$
Can someone please give me an explanation?
The motivation for the function $f(t)=t-\sqrt[3]{t+0.8}$ is that we want to create a function where a solution to the equation corresponds to a zero of the function so that we can directly use Intermediate Value theorem after evaluating the function at the 2 points $1.2, 1.3$.
You can also use the function $f(t)=\sqrt[3]{t+0.8}$. Evaluating at the 2 points $1.2$ and $1.3$ give $f(1.2)=\sqrt[3]{2}>1.2, f(1.3)=\sqrt[3]{2.1}<1.3$. You still need to use the fact that both $t$ and $f(t)$ are continuous to show that there is an intersection, which requires you to find the difference between $t$ and $f(t)$ (which would be $t-f(t)$, your original function) before you can use Intermediate Value Theorem, making this a rather roundabout way.