# Thread: intermediate value theorem question

1. ## intermediate value theorem question

I have a term test coming Nov 5th, so my posts will be here for the next couple weeks while I cram to complete my homework. Here goes:

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

This is what I have so far:

make the function equal 1000, so,

1000=(c^2)+10sin(c)

I would like to solve for c now, but for some reason, the idea in the SSM says that the IVT applies here. I would like to just start plugging in trial numbers to see what values I can get slightly above and below 1000, but I'm not even too sure on how to evaluate (let's say) 10sin(20) without a calculator.

How do I go about continuing to solve this not knowing how to evaluate the sin of my trial numbers without a calculator?

2. You're on the right track!

what you need to do is to move the 1000 to the RHS to get an equation to equal 0.

then let f(x)=(x^2)+10sinx-1000

Then all you need to do is to find a value of x for when f(x)<0, and a value of x for when f(x)>0 (you can find "nice" values of x to make the sin function easy to calculate; try x=0 for starters) - proving that there is a solution between these two points, hence this is where the IVT comes in.

3. ## Evaluating sin

Hmm.. thank you but I'm still a little confused about evaluating the sin of integers.

Sin functions tend to use radians, so would the values of c that are substituted in after I try 0 be pi values, or integers?

Like, should I try integers, such as 2 next, then 3 and so on? How do I evaluate sin(2)?
If c=2, then
0=2^2+10*sin(2)-1000
0=4+10*sin(2)-1000
((1000-4)/10)=sin(2)
498/5 = sin(2)
Nothing here tells me that I'm getting anywhere with the idea that there is an x that yields a y value of 1000.

Or do I try pi then 2*pi and so on?
If c=pi, then
0 = pi^2 + 10sin(pi) - 1000
0 = pi^2 + 10*(0) - 1000
0 = pi^2 - 1000
+-root(1000) = pi
+-10 = pi

I'm still waiting for that smile and sigh of relief... What am I doing wrong?

4. Originally Posted by krystaline86
I have a term test coming Nov 5th, so my posts will be here for the next couple weeks while I cram to complete my homework. Here goes:

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

[snip]
f(x) is continuous over [0, 100].

f(0) = 0.

f(100) = 10,000 + (a number between -10 and 10) > 1000.

Therefore by the IVT there exists a number c such that 0 < c < 100 and f(c) = 1000.

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# x^2 10 sin x

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