Verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval. Then find all numbers c that satisfy the conclusion of Rolle's Theorem.
f(x)= x sqrt(x+21) , [-21,0]
If there is more than one solution separate your answers with commas.
c =
Do you know what Rolle's Theorem says?
Rolle's theorem states that if a function f is continuous on a closed interval [a,b] and f(a) = f(b) then
f'(c) = 0 where a <= c <+ b
it is easy to see that f(-21) = -21√0 = 0
and f(0) = 0
f'(x) = (x/2+1)(x+21)^(-1/2)
setting this equal to zero and solving I got x=-2, wich lies between -21 and 0
To verify that the function satisfies the three hypotheses of Rolle's Theorem, we need to check the following:
1. The function f(x) is continuous on the closed interval [-21, 0].
2. The function f(x) is differentiable on the open interval (-21, 0).
3. The function f(a) = f(b), where a = -21 and b = 0.
1. Continuity: The function f(x) = x√(x+21) is a composition of continuous functions, namely x and √(x+21), which are both continuous on their respective domains. Therefore, f(x) is continuous on the entire interval [-21, 0].
2. Differentiability: The function f(x) is differentiable on the open interval (-21, 0) since it is composed of continuous and differentiable functions.
3. Equal values: We need to check if f(a) = f(b), where a = -21 and b = 0.
f(-21) = (-21)√(-21+21) = 0
f(0) = 0√(0+21) = 0
Since f(a) = f(b) = 0, the third hypothesis is satisfied.
Now, to find the numbers c that satisfy the conclusion of Rolle's Theorem, we need to find where f'(x) = 0 on the interval (-21, 0).
Differentiating the function f(x), we have:
f'(x) = (x/2 + 1)(x+21)^(-1/2)
Setting f'(x) equal to zero, we can solve for x:
(x/2 + 1)(x+21)^(-1/2) = 0
This equation is satisfied when either:
x/2 + 1 = 0 OR (x+21)^(-1/2) = 0
From the first solution, x = -2.
Therefore, the number c = -2 satisfies the conclusion of Rolle's Theorem.
Thus, the answer is c = -2.