veleumnihryz

2022-05-02

suppose we have following function

$f(x)={x}^{2}+10\mathrm{sin}(x)$

we should show that there exist number $c$ such that $f(c)=1000$

clearly we can solve this problem using intermediate value theorem for instance

$f(0)=0$

$f(90)={90}^{2}+10\mathrm{sin}(90)=8100+10\ast 1=8110$

and because 1000 is between these two number we can see that there exist such c so that $f(c)=1000$

am i right? thanks in advance

$f(x)={x}^{2}+10\mathrm{sin}(x)$

we should show that there exist number $c$ such that $f(c)=1000$

clearly we can solve this problem using intermediate value theorem for instance

$f(0)=0$

$f(90)={90}^{2}+10\mathrm{sin}(90)=8100+10\ast 1=8110$

and because 1000 is between these two number we can see that there exist such c so that $f(c)=1000$

am i right? thanks in advance

Trey Harrington

Beginner2022-05-03Added 10 answers

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

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