Solutions

i) Notice that solving for

$$x^{179}+\frac{163}{1+x^2+\sin^2 x}=119$$

is equivalent to solving for

$$x^{179}+\frac{163}{1+x^2+\sin^2 x}-119=0.$$

Letting the left-hand side be $f(x)$, we have

$$f(0)=163-119=44$$

and

$$f(-1)=-120+\frac{163}{2+\sin^2(-1)}\le -120+\frac{163}{2}<0.$$

Hence, by the IVT, there is a number $x$ in $[-1,0]$ such that $f(x)=0$.

ii) We are solving $f(x)=\sin x-x+1=0$. We have $f(0)=1$ and

$$f(2)=\sin 2-2+1=\sin 2-1<0.$$

Hence, by the IVT, there exists $x\in[0,2]$ such that $f(x)=0$.