Find the relative maximum and relative minimum of the graph of the function, f(x) = x^3 + 4x^2 - 3x. Determine where the relative maximum occurs and round each coordinate to the nearest hundredth. Show your work and choose the correct answer below.
(0.33,-0.52)
(0.53,17.56)
(-3.00,18.00)
(-1.39,2.32)
To find the relative maximum and relative minimum of a function, we can first find the critical points by setting the derivative equal to zero.
f(x) = x^3 + 4x^2 - 3x
f'(x) = 3x^2 + 8x - 3
Setting f'(x) = 0:
3x^2 + 8x - 3 = 0
We can solve this equation by factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in a = 3, b = 8, c = -3:
x = (-8 ± √(8^2 - 4(3)(-3))) / (2(3))
x = (-8 ± √(64 + 36)) / 6
x = (-8 ± √100) / 6
x = (-8 ± 10) / 6
This gives us two solutions:
x = (-8 + 10) / 6 = 2/6 = 1/3 ≈ 0.33
x = (-8 - 10) / 6 = -18 / 6 = -3
Now we need to determine if these critical points correspond to a relative maximum or relative minimum. We can do this by examining the second derivative of the function.
f'(x) = 3x^2 + 8x - 3
f''(x) = 6x + 8
Plugging in x = 0.33 and x = -3 into f''(x):
f''(0.33) = 6(0.33) + 8 ≈ 10.98
f''(-3) = 6(-3) + 8 = -10
Since f''(0.33) is positive, it means the function is concave up and the point (0.33, f(0.33)) is a relative minimum.
Since f''(-3) is negative, it means the function is concave down and the point (-3, f(-3)) is a relative maximum.
Now let's find the y-values of these relative extrema:
f(0.33) ≈ (0.33)^3 + 4(0.33)^2 - 3(0.33) ≈ -0.52
f(-3) = (-3)^3 + 4(-3)^2 - 3(-3) = -27 + 36 + 9 = 18
Rounding each coordinate to the nearest hundredth:
Relative minimum: (0.33, -0.52)
Relative maximum: (-3.00, 18.00)
Therefore, the correct answer is (-3.00, 18.00).
To find the relative maximum and relative minimum of a function, we need to first find the critical points. These are the points where the derivative of the function is equal to zero or where it does not exist.
Let's start by finding the derivative of the function f(x) = x^3 + 4x^2 - 3x:
f'(x) = 3x^2 + 8x - 3
To find the critical points, we set the derivative equal to zero and solve for x:
3x^2 + 8x - 3 = 0
We can use the quadratic formula to solve for x:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case, a = 3, b = 8, and c = -3. Plugging these values into the quadratic formula, we get:
x = (-8 ± sqrt(8^2 - 4 * 3 * -3)) / (2 * 3)
x = (-8 ± sqrt(64 + 36)) / 6
x = (-8 ± sqrt(100)) / 6
x = (-8 ± 10) / 6
Solving for x, we have two critical points:
x1 = (-8 + 10) / 6 = 2/3 ≈ 0.33
x2 = (-8 - 10) / 6 = -18/6 = -3
To find the y-coordinate at each critical point, we substitute these values back into the original function:
f(0.33) = (0.33)^3 + 4(0.33)^2 - 3(0.33) ≈ -0.52
f(-3) = (-3)^3 + 4(-3)^2 - 3(-3) = -27 + 36 + 9 = 18
The coordinates of the critical points are approximately (0.33, -0.52) and (-3, 18).
Now we need to determine whether each critical point is a relative maximum or minimum. To do this, we can analyze the concavity of the function.
The second derivative of f(x) is the derivative of the first derivative:
f''(x) = 6x + 8
To determine the concavity, we evaluate the second derivative at each critical point:
f''(0.33) = 6(0.33) + 8 ≈ 10
f''(-3) = 6(-3) + 8 = -10
Since f''(0.33) > 0, the point (0.33, -0.52) is a relative minimum.
Since f''(-3) < 0, the point (-3, 18) is a relative maximum.
So, the correct answer is (-3.00, 18.00).
To find the relative maximum and relative minimum of a function, we need to analyze the critical points and the behavior of the function around those points.
1. Firstly, let's find the critical points of the function. Critical points occur when the derivative of the function is equal to zero or does not exist.
To find the derivative of the function, f(x) = x^3 + 4x^2 - 3x, we differentiate it with respect to x:
f'(x) = 3x^2 + 8x - 3
Now, we set this derivative equal to zero and solve for x to find the potential critical points:
3x^2 + 8x - 3 = 0
We can solve this quadratic equation by factoring, completing the square, or by using the quadratic formula. In this case, it is easiest to use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values a = 3, b = 8, and c = -3 into the quadratic formula, we get:
x = (-8 ± √(8^2 - 4(3)(-3))) / (2(3))
x = (-8 ± √(64 + 36)) / 6
x = (-8 ± √100) / 6
x = (-8 ± 10) / 6
Simplifying further, we get two potential critical points:
x = (-8 + 10) / 6 = 2/3 ≈ 0.33
x = (-8 - 10) / 6 = -18/6 = -3
Therefore, the potential critical points are x = 0.33 and x = -3.
2. To determine if these points are relative maximum or minimum, we can use the second derivative test. Thus, we need to find the second derivative of the function.
Taking the derivative of f'(x) = 3x^2 + 8x - 3, we get:
f''(x) = 6x + 8
3. Now, we evaluate the second derivative at the potential critical points.
For the point x = 0.33, we have:
f''(0.33) = 6(0.33) + 8 = 2.98
For the point x = -3, we have:
f''(-3) = 6(-3) + 8 = -10
Since f''(0.33) > 0 and f''(-3) < 0, we can conclude the following:
a. At x = 0.33, the second derivative is positive, which means the function is concave up. This implies that x = 0.33 is a relative minimum.
b. At x = -3, the second derivative is negative, which means the function is concave down. This implies that x = -3 is a relative maximum.
4. To determine the y-coordinates of the relative maximum and minimum, we substitute these x-values back into the original function, f(x) = x^3 + 4x^2 - 3x.
For the relative maximum at x = -3:
f(-3) = (-3)^3 + 4(-3)^2 - 3(-3) = -27 + 36 + 9 = 18
For the relative minimum at x = 0.33:
f(0.33) = (0.33)^3 + 4(0.33)^2 - 3(0.33) ≈ 0.53
Therefore, the relative maximum occurs at (-3.00, 18.00), and the relative minimum occurs at (0.33, -0.52).
Comparing with the given answer choices, the correct answer is (-3.00, 18.00) and (0.33, -0.52).