Question
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?
Answers
oobleck
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.
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.
Alice
Thank you! However, wouldn't there be three values for c?
oobleck
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)