f(x) = 5x√(x+2) is continuous and differentiable on [0,2], so the MVT applies.
f(2) = 20
f(0) = 0
So, the slope of the secant is 10
MVT says that f'(c) = 20 somewhere in [0,2]
f'(x) = 5(3x+4) / 2√(x+2)
f'(c) = 0 if
5(3c+4) / 2√(c+2) = 10
5(3c+4) = 20√(c+2)
(3c+4)^2 = 16(c+2)
9c^2 + 24c + 16 = 16c+32
9c^2 + 8c - 16 = 0
c = 0.96101
That is in the interval [0,2], as predicted by the MVT.
The function f(x) = 5x sqrt x+2 satisfies the hypotheses of the Mean Value Theorem on the interval [0,2]. Find all values of c that satisfy the conclusion of the theorem.
How would you use the MVT? I tried taking the derivative, in which resulted in 5sqrtx+2 + (5x/2sqrtx+2) and by using the MVT, f'(c) = f(b)-f(a)/b-a, in which was f(2)-f(0)/2-0.
Now, I have to plug in the values...but I am not sure how you would do that and find the values of c. Please help?
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