since f(x) < 0 in the interval, we can use
f(x) = -(x^2-4)(x^2+2)
f(1) = 9
f(0) = 8
the slope of the secant is this (9-8)/(1-0) = 1
By inspection of the graph, there appear to be two values that work. But to be sure,
f'(x) = -4x(x^2-1)
So, we want to find c such that
-4c(c^2-1) = 1
solve that using your favorite tools and there are indeed two values for c.
If f(x) = |(x^2 − 4)(x^2 + 2)|, how many numbers in the interval [0, 1] satisfy the conclusion of the Mean Value Theorem?
3 answers
Thank you! However, wouldn't there be three values for c?
yes, but one is not included in the interval [0,1]
Take a look at the graph here:
https://www.wolframalpha.com/input/?i=-(x%5E2+%E2%88%92+4)(x%5E2+%2B+2)
Take a look at the graph here:
https://www.wolframalpha.com/input/?i=-(x%5E2+%E2%88%92+4)(x%5E2+%2B+2)