This is an odd function so its integral from -1 to +1 will be zero but anyway
find average value on that interval
that is the integral /(1 - -1)= integral/2
integral = (1/4)x^4 - 8x^2 from -1 to + 1
= 1/4-8 -(1/4)+8 = 0
so where is the function zero?
at x = 0 of course
I think you have misstated the problem. You seem to be doing the mean value theorem for derivatives, not integrals
If that is the case then we want
at +1, f(x) = 1-16 = -15
at -1, f(x) = -1+16 = +15
change in f = -30
change in x = 2
so average dy/dx = -15
where is derivative = -15?
f'(x) = -15 somewhere on interval
f'(c) = 3x^2 -16 = -15 (that is third on your list)
3 x^2 = 1
x = +/- sqrt(1/3) = +/- (1/3)sqrt 3
Determine if the Mean Value Theorem for Integrals applies to the function f(x) = x^3 − 16x on the interval [−1, 1]. If so, find the x-coordinates of the point(s) guaranteed to exist by the theorem.
f(a)=15
f(b)=-15
f'(x)=3x^2-16
f'(c)=-15
I am now lost from here, I do not know how to simplify -15=3c^2+16. I used a calculator to check my work and it gave me ±((√3)/3) yet when I work it out, I get ±((√1)/3). Am I just doing something wrong? Any help is appreciated.
1 answer
1 answer