Consider the function f(x)=3x^2 – 5x on the interval [-4,4]. Find the average slope of the function on this interval.
By the mean value theorem, we know there exists at least one c in the open interval (-4,4) such that f’(c) is equal to this mean slope. For this problem, there are 2 values of c that work. The smaller one and the larger one are?
f(-4) = 68
f(4) = 28
average slope = (28-68)/(4+4) = -5
f'(x) = 6x-5
f'(c) = 6c - 5 = -5
c = 0
I only get one value.
To find the average slope of the function f(x) on the interval [-4, 4], we need to calculate the slope between the two endpoints.
First, we find the derivative of the function f(x) to obtain f'(x). Differentiating the function f(x) = 3x^2 - 5x, we get f'(x) = 6x - 5.
Next, we evaluate f'(x) at the two endpoints of the interval:
At x = -4, f'(-4) = 6(-4) - 5 = -24 - 5 = -29.
At x = 4, f'(4) = 6(4) - 5 = 24 - 5 = 19.
The slope between the two endpoints is given by (f(b) - f(a)) / (b - a), where a and b are the endpoints of the interval.
Using the formula, the average slope of the function on the interval [-4, 4] is:
( f(4) - f(-4) ) / ( 4 - (-4) ) = ( (3(4)^2 - 5(4)) - (3(-4)^2 - 5(-4)) ) / ( 4 - (-4) )
= ( (48 - 20) - (48 + 20) ) / 8
= ( 28 - 68 ) / 8
= -40 / 8
= -5
So, the average slope of the function on the interval [-4, 4] is -5.
Now, according to the Mean Value Theorem, this mean slope of -5 would be equal to f'(c), where c is a value in the open interval (-4, 4). Since f'(x) = 6x - 5, we can set it equal to -5 and solve for x:
6x - 5 = -5
6x = 0
x = 0
Therefore, the two values of c that work are c = 0 and c = 0.