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Question

Consider the function 𝑓(𝑥)=7−6𝑥2f(x)=7−6x2 on the interval [−3,3][−3,3]. 

(A) Find the average or mean slope of the function on this interval, i.e. 

𝑓(3)−𝑓(−3)/3−(−3) +

(B) By the Mean Value Theorem, we know there exists a 𝑐c in the open interval (−3,3)(−3,3) such that 𝑓′(𝑐)f′(c) is equal to this mean slope. For this problem, there is only one  𝑐c that works. Find it. c =  

Answer 1

The average or or mean slope of a function f(x) on an interval [a,b] is simply the slope of the line connecting the points (a, f(a)) and (b, f(b)). That is,

m = (f(b) - f(a))/(b-a)

For our f, we have

f(3) = 7 - 6*(3)² = 7 - 54 = -47

f(-3) = 7 - 6*(-3)² = 7 - 54 = -47

and so

m = (-47 - (-47))/(3-(-3)) = 0/6 = 0.

We may also notice that our function is a quadratic equation or parabola. Such a function has a slope of zero at only one location: its vertex. If we recall that the vertex of a parabola written in the form

f(x) = a(x-h)² + k is the point (h,k), we can rearrange our equation to get

7 - 6x² = -6(x-0)² + 7

to get the vertex (0,7)

Even faster, we may realize that if the average slope between two points on a parabola is zero, then the vertex's x coordinate must be halfway between them. The midpoint of -3 and 3 is 0. Plugging x = 0 into our quadratic equation gives f(0) = 7, and our vertex is again (0,7).

Finally, if we must go the route of the Mean Value Theorem proof, we can simply set the derivative of our function equal to the mean slope:

f'(c) = -12c = 0.

Solving for c gives c = 0. Plugging this into f(x) again gives the point (0,7).