Calculus questions involving intermediate theorem?

Question

How would I solve the following two questions.

Using the intermediate value theorem to show that there is a solution of the equation $\frac{sin^2x}{2}-x+1=0$ in the interval $[0,pi]$

I showed by the IVT there is a c in $[0,pi]$ give that c is zero because $0<1$ $0>-pi+1$ but I am not sure if it did this correctly.

2.My second question asked me to sketch a function that satisfies the following conditions or prove that it is impossible $f(x)$ is a continous function on $[0,2]$ its minimum value is $-3$ but it does not have a max value.

I said this is impossible because if $f(x)$ is continous on a bounded interval it must have both a max and min value.

Answer 1

Using the intermediate value theorem to show that there is a solution of the equation $\dfrac{\sin^2x}{2}-x+1=0$ in the interval $[0,\pi]$

"I showed by the IVT there is a c in $[0,\pi]$ give that c is zero because $0<1$, $0>-pi+1$ but I am not sure if it did this correctly."

Your process in answering this is just fine: just clarify the details:

Let $f(x)=\dfrac{\sin^2x}{2}−x+1.$
Show the Intermediate Value theorem does, in fact, apply on the given interval $[0,\pi]$: Then $f(0)>0$ and $f(\pi)<0$, using your computations...etc., including the details/justifications, you posted.

The second part looks just fine, as it is, as you explained exactly why it is not possible.