Find the relative maximum and relative minimum of the graph of the function.
f(x)=x^3+4x^2-3x
To find the relative maximum and relative minimum of the graph of the function f(x) = x^3 + 4x^2 - 3x, we need to find the critical points and then determine if they correspond to a maximum or minimum.
First, let's find the derivative of f(x):
f'(x) = 3x^2 + 8x - 3.
Next, let's find the critical points by setting the derivative equal to 0:
3x^2 + 8x - 3 = 0.
This equation can be factored as (3x - 1)(x + 3) = 0.
Setting each factor equal to 0, we find two critical points:
3x - 1 = 0 -> x = 1/3,
x + 3 = 0 -> x = -3.
To determine if these critical points correspond to a maximum or minimum, we can use the second derivative test. Let's find the second derivative of f(x):
f''(x) = 6x + 8.
To apply the second derivative test, we substitute each critical point into the second derivative:
For the critical point x = 1/3:
f''(1/3) = 6(1/3) + 8 = 2 + 8 = 10.
Since f''(1/3) is positive, this means that the function has a relative minimum at x = 1/3.
For the critical point x = -3:
f''(-3) = 6(-3) + 8 = -18 + 8 = -10.
Since f''(-3) is negative, this means that the function has a relative maximum at x = -3.
Therefore, the relative minimum is at (1/3, f(1/3)) and the relative maximum is at (-3, f(-3)).
To find the relative maximum and relative minimum of the function f(x) = x^3 + 4x^2 - 3x, we need to find the critical points by taking the derivative and setting it equal to 0.
Step 1: Find the derivative of f(x):
f'(x) = 3x^2 + 8x - 3
Step 2: Set the derivative equal to 0 and solve for x:
3x^2 + 8x - 3 = 0
Unfortunately, the equation does not factor nicely, so we will need to use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 3, b = 8, and c = -3:
x = (-8 ± √(8^2 - 4(3)(-3))) / (2(3))
x = (-8 ± √(64 + 36)) / 6
x = (-8 ± √100) / 6
x = (-8 ± 10) / 6
The two possible solutions are x = (-8 + 10) / 6 = 1/3 and x = (-8 - 10) / 6 = -3.
Step 3: Test the critical points to determine if they are relative maximum or minimum:
To do this, we can use the second derivative test. Find the second derivative of f(x):
f''(x) = 6x + 8
For x = 1/3, evaluate f''(1/3):
f''(1/3) = 6(1/3) + 8
f''(1/3) = 2 + 8
f''(1/3) = 10
Since the second derivative is positive at x = 1/3, it means that f(x) has a relative minimum at x = 1/3.
For x = -3, evaluate f''(-3):
f''(-3) = 6(-3) + 8
f''(-3) = -18 + 8
f''(-3) = -10
Since the second derivative is negative at x = -3, it means that f(x) has a relative maximum at x = -3.
Step 4: Find the corresponding y-values for the relative maximum and relative minimum:
To find the y-values, substitute the x-values into the original function f(x):
For x = 1/3:
f(1/3) = (1/3)^3 + 4(1/3)^2 - 3(1/3)
f(1/3) = 1/27 + 4/9 - 1
f(1/3) = 1/27 + 12/27 - 27/27
f(1/3) = -14/27
So, the relative minimum is at (1/3, -14/27).
For x = -3:
f(-3) = (-3)^3 + 4(-3)^2 - 3(-3)
f(-3) = -27 + 36 + 9
f(-3) = 18
So, the relative maximum is at (-3, 18).
Therefore, the relative maximum is at (-3, 18) and the relative minimum is at (1/3, -14/27).