# Thread: show that there is a number c such that f(c) = 1000

1. ## show that there is a number c such that f(c) = 1000

Question: If f(x) = x^2 + 10 sinx, show that there is a number c such that f(c) = 1000.

I'm not too sure I understand the question but I plugged in 1000 as the y value in that rule and don't really no how to solve for x.

Any help would be greatly appreciated!

2. Originally Posted by s3a
Question: If f(x) = x^2 + 10 sinx, show that there is a number c such that f(c) = 1000.

I'm not too sure I understand the question but I plugged in 1000 as the y value in that rule and don't really no how to solve for x.

Any help would be greatly appreciated!
$f(31) \approx 957$

$f(32) \approx 1030$

what does the Intermediate Value Theorem tell you?

3. That there is a point (c,N) in between (31, 957) and (32, 1030)?

4. Originally Posted by s3a
Question: If f(x) = x^2 + 10 sinx, show that there is a number c such that f(c) = 1000.

I'm not too sure I understand the question but I plugged in 1000 as the y value in that rule and don't really no how to solve for x.

Any help would be greatly appreciated!
$f(0)=0$, $f(\infty)=\lim\nolimits_{x\to\infty}f(x)=\infty$, because of the continuity of $f$ and the intermediate value theorem, there exists $x_{0}\geq0$ such that $f(x_{0})=y_{0}$ for any fixed $y_{0}\geq0$.