Find the values of x that give relative extrema for the function f(x)=3x^5-5x^3
A. Relative maximum: x= 1; Relative minimum: x=sqrt(5/3)
B. Relative maximum: x=-1; Relative minimum: x=1
C. Relative maxima: x=+or- 1; Relative minimum: x=0
D. Relative maximum: x=0; Relative maxima: x=+or- 1
E. none of these
Solve without a graph
f ' (x) = 15x^4 - 15x^2
= 0 for a max/min
15x^2(x^2 - 1) = 0
x=0 or x=±1
to determine if these yield a max or a min look at the 2nd derivative
f '' (x) = 60x^3 - 30x
f "(0) = 0 , so x = 0 yields a point of inflection
f "(1) = 60-30 > 0 , so x = 1 yields a minimum
f "(-1) = -60 -(-30) < 0 , so x = -1 yields a maximum
I think B fits my conclusions
confirmation:
http://www.wolframalpha.com/input/?i=plot+y+%3D3x%5E5-5x%5E3
Well, finding relative extrema involves finding where the derivative of the function is equal to zero. Let's start by finding the derivative of f(x).
f(x) = 3x^5 - 5x^3
f'(x) = 15x^4 - 15x^2
Now, let's set the derivative equal to zero and solve for x:
15x^4 - 15x^2 = 0
Factoring out 15x^2, we get:
15x^2(x^2 - 1) = 0
Considering the two separate factors:
1) 15x^2 = 0
This gives us x = 0 as a potential critical point.
2) x^2 - 1 = 0
Using the difference of squares, we have (x - 1)(x + 1) = 0
This gives us two more potential critical points: x = 1 and x = -1.
Now we have three potential critical points: x = 0, x = 1, and x = -1.
To determine whether these points are relative extrema, we can use the second derivative test. However, computing the second derivative would unnecessarily complicate things here.
Instead, we can plug in these values into our original function and compare the values.
f(0) = 3(0)^5 - 5(0)^3 = 0
f(1) = 3(1)^5 - 5(1)^3 = -2
f(-1) = 3(-1)^5 - 5(-1)^3 = -2
We see that f(1) and f(-1) have the same value of -2, which means they are points of relative minimum.
Therefore, the correct answer is:
B. Relative maximum: x = -1; Relative minimum: x = 1
To find the values of x that give relative extrema for the function f(x) = 3x^5 - 5x^3, we need to find the critical points and then determine whether they correspond to relative maximum or minimum points.
Step 1: Find the derivative of f(x)
The derivative of f(x) = 3x^5 - 5x^3 can be found by applying the power rule of derivatives. The derivative is given by f'(x) = 15x^4 - 15x^2.
Step 2: Set the derivative equal to zero to find the critical points
To find the critical points, set f'(x) = 0 and solve for x.
15x^4 - 15x^2 = 0
Factoring out 15x^2, we get:
15x^2(x^2 - 1) = 0
Setting each factor equal to zero, we have:
15x^2 = 0 --> x = 0
x^2 - 1 = 0 --> x^2 = 1 --> x = ±1
So, the critical points are x = 0 and x = ±1.
Step 3: Determine the nature of the critical points
To determine whether each critical point corresponds to a relative maximum or minimum, we can test the sign of the derivative around each point.
Taking values to the left and right of x = 0, let's consider x = -1 and x = 1 as test points.
If x = -1, we have f'(-1) = 15(-1)^4 - 15(-1)^2 = 15 - 15 = 0.
Since the derivative changes sign from positive to negative, x = -1 corresponds to a relative maximum.
If x = 1, we have f'(1) = 15(1)^4 - 15(1)^2 = 15 - 15 = 0.
Since the derivative changes sign from negative to positive, x = 1 corresponds to a relative minimum.
Therefore, the values of x that give relative extrema for the function f(x) = 3x^5 - 5x^3 are:
- Relative maximum: x = -1
- Relative minimum: x = 1
So, the correct answer is option B: Relative maximum: x = -1; Relative minimum: x = 1.
To find the values of x that give relative extrema for the function f(x) = 3x^5 - 5x^3, we need to find the critical points where the derivative of the function is equal to zero or undefined. These critical points can correspond to relative extrema.
1. Take the derivative of f(x) with respect to x:
f'(x) = 15x^4 - 15x^2
2. Set f'(x) equal to zero to find the critical points:
15x^4 - 15x^2 = 0
3. Factor out 15x^2 from the equation:
15x^2(x^2 - 1) = 0
4. Apply the zero product property:
15x^2 = 0 or x^2 - 1 = 0
5. Solve each equation separately:
For 15x^2 = 0, we get x = 0.
For x^2 - 1 = 0, we get x = ±1.
These values of x give us the critical points.
6. To determine whether each critical point is a relative maximum or minimum, we can use the second derivative test.
Take the second derivative of f(x) with respect to x:
f''(x) = 60x^3 - 30x
7. Plug in the values of the critical points into f''(x) to determine their nature:
For x = 0:
f''(0) = 60(0)^3 - 30(0) = 0
Since the second derivative is zero, the test is inconclusive.
For x = 1:
f''(1) = 60(1)^3 - 30(1) = 30
Since the second derivative is positive, the point (1, f(1)) is a relative minimum.
For x = -1:
f''(-1) = 60(-1)^3 - 30(-1) = -30
Since the second derivative is negative, the point (-1, f(-1)) is a relative maximum.
Therefore, the values of x that give relative extrema for the function f(x) = 3x^5 - 5x^3 are:
Relative maximum: x = -1
Relative minimum: x = 1
So, the correct answer is B. Relative maximum: x = -1; Relative minimum: x = 1.