Consider the function
f(x)=–2x^3–3x^2+2x–2
Find the average slope of this function on the interval (2,6). ?
By the Mean Value Theorem, we know there exists a c in the open interval (2,6) such that f'(c) is equal to this mean slope. Find the value of c in the interval which works ?
To find the average slope of the function on the interval (2,6), we need to find the derivative of the function and then evaluate it at some point within that interval.
Step 1: Find the derivative of the function f(x) by applying the power rule for derivatives.
f'(x) = -6x^2 - 6x + 2
Step 2: To find the average slope, we need to calculate the difference in the values of the function at the endpoints of the interval and divide it by the width of the interval.
Let's calculate the difference in the values of the function at the endpoints:
f(6) - f(2) = [-2(6)^3 - 3(6)^2 + 2(6) - 2] - [-2(2)^3 - 3(2)^2 + 2(2) - 2]
= [-432 - 108 + 12 - 2] - [-16 - 12 + 4 - 2]
= [-530] - [-22]
= -508
Now, let's calculate the width of the interval:
6 - 2 = 4
Step 3: Finally, divide the difference in function values by the width of the interval to find the average slope:
Average slope = (-508) / 4
= -127
Now, we need to find a value of c in the open interval (2,6) such that f'(c) is equal to this mean slope (-127).
To find this value, set f'(c) = -127 and solve for c:
-6c^2 - 6c + 2 = -127
Simplifying the equation further, we get:
6c^2 + 6c - 129 = 0
Now, you can solve this quadratic equation using methods such as factoring, completing the square, or using the quadratic formula to find the value of c.