Calculus IVT and Discontinuity?

Question

Hi. Can someone help me with these calculus questions?

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

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval.

cosx = x (0, 1)

I am so lost with these questions. Can someone explain them? Thank you.

Answer 1

IVT is one of those dumb existence theorems that looks hard, but is incredibly simple.

For the first one...

f(x) = x^2 + 10sinx

by substitution f(100) = 9995 (roughly)

and f(101) = 10205 (roughly)

now if, in fact, f(x) is CONTINUOUS (and it is) then there must be another value I can substitute in to get 1000. There has to be! (you didn't have to choose 100 and 101)

f(100) is below 1000 and f(101) is above 1000. There has to be a number c in between 100 and 101 such that f(c) = 1000

same thing for number 2

cos x = x or cos x -x = 0

let's try x = 0

cos (0) - 0 = 1 (make sure you're in radians)

how bout 1

cos 1 - 1 = -.46 (roughly)

notice when I plugged in 0 the sign is positive. when I plugged in 1 the sign is negative. That means that there must be a root between 0 and 1! The graph had to have crossed over the x-axis at some point between 0 and 1. I don't care exactly where, but I know it had to have crossed because of the sign change. It's kinda common sense.