# suppose we have following function f <mo stretchy="false">( x <mo stretchy="false">)

suppose we have following function
$f\left(x\right)={x}^{2}+10\mathrm{sin}\left(x\right)$
we should show that there exist number $c$ such that $f\left(c\right)=1000$

clearly we can solve this problem using intermediate value theorem for instance
$f\left(0\right)=0$
$f\left(90\right)={90}^{2}+10\mathrm{sin}\left(90\right)=8100+10\ast 1=8110$
and because 1000 is between these two number we can see that there exist such c so that $f\left(c\right)=1000$

am i right? thanks in advance
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Trey Harrington
${f}^{\prime }\left(x\right)=2x+10\mathrm{cos}x$ is clearly positive if $x\ge 6$, hence our function is increasing over the interval $\left[6,33\right]$ and since 1000 is between $f\left(6\right)$ and $f\left(33\right)$, there is some (only one) $c\in \left(6,33\right)$ such that $f\left(c\right)=1000$.