Find the average or mean slope of 2x^3 - 6x^2 - 90x +3 on the interval [-6,10]?
By the Mean Value Theorem, we know there exists at least one c in the open interval ( -6 , 10 ) such that f'( c) is equal to this mean slope.
::>> I already found the mean slope of the function which is 38. Now I'm trying to find "c". I can't get c.
6x^2 - 12 x -90 = 38 ??? assuming your 38 is correct
6 x^2 - 12 x - 128 = 0
3x^2 - 6 x - 64 = 0
x =5.73
To find the value of "c" that satisfies the Mean Value Theorem, you need to find where the derivative of the function is equal to the mean slope. In this case, the mean slope is given as 38.
First, let's find the derivative of the function:
f(x) = 2x^3 - 6x^2 - 90x + 3
Using the power rule, we can differentiate each term of the function:
f'(x) = (3 * 2) x^(3-1) + (2 * -6) x^(2-1) + (-90)
Simplifying gives:
f'(x) = 6x^2 - 12x - 90
Now, set this derivative equal to the mean slope and solve for "x":
6x^2 - 12x - 90 = 38
Rearrange the equation:
6x^2 - 12x - 90 - 38 = 0
Combine like terms:
6x^2 - 12x - 128 = 0
Now, you can use the quadratic formula to solve for "x":
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values for a, b, and c:
a = 6
b = -12
c = -128
x = (-(-12) ± √((-12)^2 - 4 * 6 * (-128))) / (2 * 6)
Simplifying further gives:
x = (12 ± √(144 + 3072)) / 12
x = (12 ± √(3216)) / 12
x = (12 ± 56.75) / 12
This gives two potential values for "c":
c₁ = (12 + 56.75) / 12 = 4.98
c₂ = (12 - 56.75) / 12 = -3.55
Therefore, there are two values of "c" within the interval [-6, 10] that satisfy the Mean Value Theorem: c₁ ≈ 4.98 and c₂ ≈ -3.55.