Single Variable Calculus, Chapter 2, 2.5, Section 2.5, Problem 45

Suppose $f(x) = x^2 + 10 \sin x$, show that there is a number $c$ such that $f(c) = 1000$.

Let $g(x) = x^2 + 10 \sin x - 1000$
Based from the definition of Intermediate value Theorem,
There exist a solution $c$ for the function between the interval $(a,b)$ suppose that the function is continuous
on the given interval. So we take $a$ and $b$ to be 0 and 50 respectively and assume the function $g(x)$
is continuous on the interval (0,50). So we have,


$
\begin{equation}
\begin{aligned}
g(0) & = (0)^2 + 10 \sin (0) - 1000 = -1000\\
g(50)& = (50)^2 + 10 \sin (50) - 1000 = 1497.3763
\end{aligned}
\end{equation}
$


By using Intermediate Value Theorem. We prove that...

$
\begin{equation}
\begin{aligned}
& \text{if } 0 < c < 50 && \text{then } \quad g(0) < g(c) < g(50)\\
& \text{So, }\\
& \text{if } 0 < c < 50 && \text{then } \quad -1000 < 1000 < 1497.3763
\end{aligned}
\end{equation}
$

Therefore,
There exist a such solution $c$ for $f(x) = x^2 + 10 \sin x$