1. Find all relative extrema of the function f(x)=x^(6/7)-3. Use the Second Derivative Test where applicable.
2. Find all relative extrema of the function f(x)=2x^4-16x^3+4. Use the Second Derivative Test where applicable.
To find the relative extrema of a function, we need to find the critical points where the derivative is either zero or undefined and then apply the Second Derivative Test.
1. For the function f(x) = x^(6/7) - 3:
a. Find the first derivative f'(x):
f'(x) = (6/7)x^(-1/7)
b. Set the derivative equal to zero and solve for x to find the critical point:
(6/7)x^(-1/7) = 0
x^(-1/7) = 0
Since x^(-1/7) cannot equal zero, there are no critical points where the derivative is zero.
c. Determine the points where the derivative is undefined:
The derivative is defined for all real values of x.
d. Apply the Second Derivative Test to the critical points found:
Since there are no critical points, we can skip this step.
e. Conclusion:
There are no relative extrema for the function f(x) = x^(6/7) - 3.
2. For the function f(x) = 2x^4 - 16x^3 + 4:
a. Find the first derivative f'(x):
f'(x) = 8x^3 - 48x^2
b. Set the derivative equal to zero and solve for x to find the critical points:
8x^3 - 48x^2 = 0
8x^2(x - 6) = 0
x = 0 or x = 6
c. Determine the points where the derivative is undefined:
The derivative is defined for all real values of x.
d. Apply the Second Derivative Test to the critical points found:
i. Find the second derivative f''(x):
f''(x) = 24x^2 - 96x
ii. Substitute the critical points into the second derivative:
For x = 0: f''(0) = 0
For x = 6: f''(6) = 144
iii. Analyze the results:
- When the second derivative is zero or undefined (0 or negative in this case), the Second Derivative Test is inconclusive.
- When the second derivative is positive (greater than zero), there is a relative minimum.
- Since f''(0) is zero, we cannot determine the nature of the critical point at x = 0 using the Second Derivative Test.
- Since f''(6) is positive, there is a relative minimum at x = 6.
e. Conclusion:
The function f(x) = 2x^4 - 16x^3 + 4 has a relative minimum at x = 6. There are no other relative extrema.
as you know,
extrema where f'=0 and f''≠0
1.
f' = 6/7 x^(-1/7) is never zero
2.
f' = 24x^2 - 96x = 24x(x-4)
f'' = 48x
should be easy now to locate extrema