Question

please help, if f(x)=x^2+10sin x , show that there is a number C such f(c)=1000

Answer 1

C^2+10sin C=1000

Answer 2

Start off with an arbitrary number that you know will get a result lower than 1000. We'll call it a.

We'll say that a =0 So f(a)=0^2+10sin(0)=0

Next we'll pick a number that is going to give a result higher than 1000 that we'll call b.

We'll pick 21pi/2=b. Why did I pick this? No particular reason other than I know it will be higher than 1000.

So b=21pi/2 So f(b)=(21pi/2)^2 +10 sin(21pi/2)=some number larger than 1000.

Because the equation is continuous and differentiable, and there are an a and b such that a<c<b, there must be a c such that f(c)=1000.

Answer 3

@whoatethecheese has used "the intermediate value theorem" above
just in case you are curious...