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.
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
To determine if the Mean Value Theorem for Integrals applies to a function on a given interval, we need to check two conditions:
1. The function must be continuous on the closed interval [a, b].
2. The function must be differentiable on the open interval (a, b).
Let's go through these conditions to determine if the Mean Value Theorem applies to the function f(x) = x^3 − 16x on the interval [−1, 1].
1. Continuity:
To check continuity, we need to ensure that the function is continuous on the closed interval [−1, 1]. The function f(x) = x^3 − 16x is a polynomial, and polynomials are continuous on their entire domain. So, f(x) = x^3 − 16x is continuous on the interval [−1, 1].
2. Differentiability:
To check differentiability, we need to ensure that the function is differentiable on the open interval (−1, 1). The function f(x) = x^3 − 16x is also a polynomial, and polynomials are differentiable on their entire domain. So, f(x) = x^3 − 16x is differentiable on the interval (−1, 1).
Since both conditions are satisfied, the Mean Value Theorem for Integrals does apply to the function f(x) = x^3 − 16x on the interval [−1, 1].
Now, let's find the x-coordinates of the point(s) guaranteed to exist by the theorem.
According to the Mean Value Theorem for Integrals, if a function satisfies the given conditions, then there exists at least one value 'c' in the open interval (a, b) such that f'(c) = [f(b) - f(a)] / (b - a).
Here, a = -1, b = 1, f(a) = 15, f(b) = -15, and the derivative of f(x), f'(x) = 3x^2 - 16.
Substituting these values into the equation, we get:
f'(c) = [-15 - 15] / (1 - (-1))
f'(c) = [-30] / (2)
f'(c) = -15
To find the x-coordinate(s) 'c', we need to solve the equation f'(c) = -15 for 'c'.
3x^2 - 16 = -15
3x^2 = 1
x^2 = 1/3
x = ±√(1/3)
So, the x-coordinates of the point(s) guaranteed to exist by the Mean Value Theorem are ±√(1/3).