I'm having the hardest time with this Calculus problem on Mean Value Theorem, can anyone explain the right answer to me please?
Let f(x)= 2(x^3)- 3(x^2)+ 2x- 2 on [-1, 2]. Find all numbers c,
satisfying the conclusion to the Mean Value Theorem.
Let f(x)= 2(x^3)- 3(x^2)+ 2x- 2 on [-1, 2]. Find all numbers c,
satisfying the conclusion to the Mean Value Theorem.
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First find the slope of the secant line between x = -1 and x = 2
f(2) = 16 - 12 + 4 - 2 = 6
f(-1) = -2 - 3 - 2 - 2 = -9
So your pts to compute the slope of the secant line is:
(2,6) and (-1,-9)
m = ∆y / ∆x
m = (6 + 9) / (2 + 1)
m = 15/3 = 5
f'(x) = 6x^2 - 6x + 2
The goal is to find when the slope of the secant line is equivalent to the slope of the tangent line.
5= 6x^2 - 6x + 2
0 = 6x^2 - 6x - 3
0 = 3(2x^2 - 2x - 1)
0 = 2x^2 - 2x - 1
x = 2 ± √(4 - (4*2*-1)) / 4
x = 2 ± √(4 + 8) / 4
x = 2±√(12)/4
x = 2 ±2√(3) / 4
x = (1 ± √(3)) / 4
Both fall in our allowed interval of -1 to 2
f(2) = 16 - 12 + 4 - 2 = 6
f(-1) = -2 - 3 - 2 - 2 = -9
So your pts to compute the slope of the secant line is:
(2,6) and (-1,-9)
m = ∆y / ∆x
m = (6 + 9) / (2 + 1)
m = 15/3 = 5
f'(x) = 6x^2 - 6x + 2
The goal is to find when the slope of the secant line is equivalent to the slope of the tangent line.
5= 6x^2 - 6x + 2
0 = 6x^2 - 6x - 3
0 = 3(2x^2 - 2x - 1)
0 = 2x^2 - 2x - 1
x = 2 ± √(4 - (4*2*-1)) / 4
x = 2 ± √(4 + 8) / 4
x = 2±√(12)/4
x = 2 ±2√(3) / 4
x = (1 ± √(3)) / 4
Both fall in our allowed interval of -1 to 2