Consider the function f (x)=7−6x^2 on the interval [ − 3 , 3 ] . Find the average or mean slope of the function on this interval, i.e.
f(3)-f(-3) / 3-(-3)=
By the Mean Value Theorem, we know there exists a c in the open interval (−3,3) such that f'(c) is equal to this mean slope. For this problem, there is only one c that works. Find it.
=
ok, you have the mean slope, which is 0
So, where is f'(x) = 0?
f' = -12x
so f'(0) = 0
This should all have been clear, since f(x) is a parabola, with x=0 as the axis of symmetry, and the interval [-3,3] has its center there.
To find the average or mean slope of the function f(x) on the interval [−3, 3], we need to calculate the difference in function values divided by the difference in x-values.
The difference in function values is given by f(3) - f(-3), and the difference in x-values is 3 - (-3).
Let's start by finding f(3) and f(-3):
f(3) = 7 - 6(3)^2 = 7 - 6(9) = 7 - 54 = -47
f(-3) = 7 - 6(-3)^2 = 7 - 6(9) = 7 - 54 = -47
Now, let's calculate the difference in function values:
f(3) - f(-3) = -47 - (-47) = -47 + 47 = 0
Next, let's calculate the difference in x-values:
3 - (-3) = 3 + 3 = 6
Finally, we can calculate the average or mean slope:
(f(3) - f(-3)) / (3 - (-3)) = 0 / 6 = 0
According to the Mean Value Theorem, there exists a c in the open interval (-3, 3) such that f'(c) is equal to this mean slope. Since the mean slope is 0, we need to find a value c where f'(c) equals 0.
To find this value, we need to find the derivative of the function f(x) and set it equal to 0:
f'(x) = -12x
Setting f'(x) equal to 0:
-12x = 0
x = 0
Therefore, the value c that satisfies f'(c) = 0 and lies in the open interval (-3, 3) is c = 0.