Find the relative maximum and relative minimum of the graph of the function.

f(x)=x^3+4x^2-3x

1 answer

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)).