Determine if Rolle's Theorem can be applied. If it can, find all values of c such that f'(c)=0.
f(x)=x^3-9x, [-3,3]
f(x) is continuous over the domain. [-3,3]
f(x) is differential.
f'(x) = 3x2 -9
f(x) has two equal value points on the independent axis.
f(-3) = f(3) = 0
Therefore Rolle's Theorem Applies
.: Exists c such that f'(c)=0
Solve for c:
0 = 3c2 - 9
...
Ahh! That's the answer I had but I was doubting it because my partner got something else! Thanks!
To determine if Rolle's Theorem can be applied to a function, we need to check three conditions:
1. Function Continuity: The function f(x) must be continuous on the closed interval [-3, 3].
2. Function Differentiability: The function f(x) must be differentiable on the open interval (-3, 3), except possibly at the endpoints.
3. Equal Function Values: The function f(x) must satisfy f(-3) = f(3).
Let's check these conditions one by one for the given function f(x) = x^3 - 9x, on the interval [-3, 3].
1. Function Continuity: The function f(x) = x^3 - 9x is a polynomial function, and thus it is continuous on its domain, which includes the closed interval [-3, 3]. So, the first condition is satisfied.
2. Function Differentiability: The derivative of f(x) is f'(x) = 3x^2 - 9. The derivative is a polynomial function and is defined for all values of x. Therefore, the function f(x) is differentiable on the open interval (-3, 3). This condition is also satisfied.
3. Equal Function Values: To check if f(-3) = f(3), substitute the values of -3 and 3 into the function:
f(-3) = (-3)^3 - 9(-3) = -27 + 27 = 0
f(3) = (3)^3 - 9(3) = 27 - 27 = 0
Since f(-3) = f(3) = 0, the third condition is satisfied as well.
Since all three conditions are satisfied, we can conclude that Rolle's Theorem can be applied to the function f(x) = x^3 - 9x on the interval [-3, 3].
Now, to find all values of c such that f'(c) = 0, we need to find the critical points of the function f(x). Critical points occur where the derivative is equal to zero or undefined. In this case, the derivative is defined for all x values and only equals zero at specific points.
Setting f'(x) = 0, we have:
3x^2 - 9 = 0
Now, solve this equation to find the values of x (or c):
3x^2 - 9 = 0
3x^2 = 9
x^2 = 3
x = ±√3
Therefore, the values of c such that f'(c) = 0 are c = √3 and c = -√3.
Note that these values are within the interval [-3, 3].