16) Consider the function f(x)=2x3−6x2−90x+5 on the interval [−6,7]. Find the average or mean slope of the function on this interval.
By the Mean Value Theorem, we know there exists at least one c in the open interval (−6,7) such that f′(c) is equal to this mean slope.
Find all values of c that work and list them (separated by commas) in the box below. If there are no values of c that work.
List of numbers:
To find the mean slope of a function on an interval, we need to find the average rate of change of the function over that interval.
First, let's find the derivative of the function f(x) = 2x^3 - 6x^2 - 90x + 5. Differentiating the function gives us:
f'(x) = 6x^2 - 12x - 90
Next, we can calculate the slope between the endpoints of the interval [-6, 7] using the formula:
mean slope = (f(7) - f(-6))/(7 - (-6))
Substitute the values of f(x) into the mean slope formula:
mean slope = (2(7)^3 - 6(7)^2 - 90(7) + 5 - [2(-6)^3 - 6(-6)^2 - 90(-6) + 5])/(7 - (-6))
Simplify:
mean slope = (686 - 294 - 630 + 5 - (-432 + 216 + 540 + 5))/13
mean slope = (686 - 294 - 630 + 5 + 432 - 216 - 540 - 5)/13
mean slope = (441)/13
mean slope ≈ 33.923
According to the Mean Value Theorem, there exists at least one c in the open interval (-6, 7) such that f'(c) is equal to this mean slope. To find the values of c that satisfy f'(c) = 33.923, we set the derivative f'(x) equal to the mean slope:
6c^2 - 12c - 90 = 33.923
Rearrange the equation:
6c^2 - 12c - 123.923 = 0
To solve for c, we can use the quadratic formula:
c = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = 6, b = -12, and c = -123.923. Plugging these values into the quadratic formula:
c = (-(-12) ± √((-12)^2 - 4(6)(-123.923)))/(2(6))
c = (12 ± √(144 + 1487.382))/(12)
c = (12 ± √1631.382)/(12)
Calculating the square root:
c ≈ (12 ± 40.386)/12
So, the values of c, which satisfy f'(c) = 33.923, are approximately:
c ≈ 5.365 or c ≈ -1.865
Therefore, the values of c that work are 5.365 and -1.865.