Verify that the function satisfies the three hypotheses of Rolle's Therorem
on the given interval. Then find all numbers c that satisfy the conclusiton
of Rolle's Theorem. f(x)= x*sqrt(x+6) [-6,0]
f is continuous and differential
f(-6) =-6*sqrt(-6+6) =0
f(0)=0
f(x) = x (x+6^1/2)
f'(x)= (x+6^1/2) + (1/2x +6)^1/2
0=(x+6^1/2) + (1/2x +6)^1/2
x=-2.918
I don't think this is correct so could you please tell me what I did wrong?
Thanks.
f(x)= x*sqrt(x+6)--->
f'(x) = sqrt(x+6) + x/[2sqrt(x+6)]
f'(x) = 0 --->
x = -4
To verify that a function satisfies the three hypotheses of Rolle's Theorem, the following conditions must be met:
1. The function must be continuous on the closed interval [-6,0].
2. The function must be differentiable on the open interval (-6,0).
3. The function must have equal values at the endpoints of the interval.
Let's go through each condition for the given function f(x) = x*sqrt(x+6) on the interval [-6,0]:
1. Continuity: To check for continuity, we need to evaluate f(x) at the endpoints of the interval:
f(-6) = -6 * sqrt(-6 + 6) = 0
f(0) = 0
Both f(-6) and f(0) are equal to 0, which means the function is continuous on the interval [-6,0].
2. Differentiability: To check for differentiability, we need to find the derivative of f(x):
f'(x) = sqrt(x+6) + x/[2sqrt(x+6)]
The function is differentiable except at the point where the denominator becomes zero. Let's find that point:
sqrt(x+6) + x/[2sqrt(x+6)] = 0
Simplifying the equation, we get:
sqrt(x+6) = -x/[2sqrt(x+6)]
Squaring both sides of the equation, we get:
x + 6 = x^2/[4(x+6)]
4(x + 6) = x^2
4x + 24 = x^2
x^2 - 4x - 24 = 0
(x - 6)(x + 4) = 0
So the function is not differentiable at x = 6 and x = -4.
3. Equal values at endpoints: We have already checked this in the continuity step and found that f(-6) = f(0) = 0.
Overall, the function f(x) = x*sqrt(x+6) satisfies the three hypotheses of Rolle's Theorem on the interval [-6,0].
Now, let's find the values of c that satisfy the conclusion of Rolle's Theorem. According to Rolle's Theorem, there exists at least one value c in the interval (-6,0) such that f'(c) = 0.
We have the derivative:
f'(x) = sqrt(x+6) + x/[2sqrt(x+6)]
Setting f'(x) equal to zero, we get:
sqrt(x+6) + x/[2sqrt(x+6)] = 0
Simplifying the equation, we get:
sqrt(x+6) = -x/[2sqrt(x+6)]
Squaring both sides of the equation, we get:
x + 6 = x^2/[4(x+6)]
4(x + 6) = x^2
4x + 24 = x^2
x^2 - 4x - 24 = 0
Solving this quadratic equation, we find two solutions:
x = -6 or x = 10
However, we need to find c in the interval (-6,0). Therefore, the only solution that satisfies the conclusion of Rolle's Theorem is:
c = -4