Find the relative maximum and relative minimum of the graph of the function, f(x) = x^3 + 2x^2 - 2x. Determine where the relative maximum occurs and round each coordinate to the nearest hundredth. Show your work.
(0.39,-0.42)
(-2.68,-1.65)
(-0.75,5.06)
(-1.72,4.27)
To find the relative extrema of the function, we need to find the critical points by taking the derivative of the function and setting it equal to zero.
f(x) = x^3 + 2x^2 - 2x
f'(x) = 3x^2 + 4x - 2
Setting f'(x) = 0:
3x^2 + 4x - 2 = 0
To solve for x, we can use the quadratic formula:
x = (-4 ± √(4^2 - 4(3)(-2))) / (2(3))
x = (-4 ± √(16 + 24)) / 6
x = (-4 ± √40) / 6
x = (-4 ± 2√10) / 6
x = (-2 ± √10) / 3
So the critical points are (-2 + √10) / 3 and (-2 - √10) / 3.
To determine if each critical point is a relative maximum or minimum, we can use the second derivative test. Taking the second derivative:
f''(x) = 6x + 4
Evaluating f'' at each critical point:
f''((-2 + √10) / 3) = 6(-2 + √10) / 3 + 4 = -2 + 2√10 + 4 = 2√10 + 2
f''((-2 - √10) / 3) = 6(-2 - √10) / 3 + 4 = -2 - 2√10 + 4 = -2√10 + 2
Since f''((-2 + √10) / 3) is positive, the point (-2 + √10) / 3 is a relative minimum.
Since f''((-2 - √10) / 3) is negative, the point (-2 - √10) / 3 is a relative maximum.
Now we can evaluate the function at each critical point to find the coordinates of the relative extrema:
f((-2 + √10) / 3) = ((-2 + √10) / 3)^3 + 2((-2 + √10) / 3)^2 - 2(-2 + √10) / 3
f((-2 + √10) / 3) ≈ 5.064
f((-2 - √10) / 3) = ((-2 - √10) / 3)^3 + 2((-2 - √10) / 3)^2 - 2(-2 - √10) / 3
f((-2 - √10) / 3) ≈ -1.647
Rounding to the nearest hundredth, we have the relative maximum at (-2.68, -1.65) and the relative minimum at (-0.75, 5.06). Therefore, the correct answer is:
(-2.68, -1.65)
(-0.75, 5.06)
To find the relative maximum and relative minimum of a function, we first need to find the critical points of the function. These are the points where the derivative of the function equals zero or is undefined.
Let's start by finding the derivative of the function f(x) = x^3 + 2x^2 - 2x. Taking the derivative of each term separately using the power rule:
f'(x) = 3x^2 + 4x - 2
Setting f'(x) equal to zero:
3x^2 + 4x - 2 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = 3, b = 4, and c = -2. Plugging in these values:
x = (-4 ± √(4^2 - 4(3)(-2))) / (2(3))
= (-4 ± √(16 + 24)) / 6
= (-4 ± √40) / 6
= (-4 ± 2√10) / 6
Simplifying further:
x = (-2 ± √10) / 3
Now we have the two solutions for x, which represent the critical points of the function.
1. For x = (-2 + √10) / 3:
Substituting this value back into the original function:
f((-2 + √10) / 3) = ((-2 + √10) / 3)^3 + 2((-2 + √10) / 3)^2 - 2((-2 + √10) / 3)
Using a calculator to evaluate this expression, we get an approximate value of:
f((-2 + √10) / 3) ≈ -1.65
So, the point is approximately (-2.68, -1.65).
2. For x = (-2 - √10) / 3:
Substituting this value back into the original function:
f((-2 - √10) / 3) = ((-2 - √10) / 3)^3 + 2((-2 - √10) / 3)^2 - 2((-2 - √10) / 3)
Using a calculator to evaluate this expression, we get an approximate value of:
f((-2 - √10) / 3) ≈ 5.06
So, the point is approximately (-0.75, 5.06).
Therefore, the relative maximum occurs at approximately (-2.68, -1.65), and the relative minimum occurs at approximately (-0.75, 5.06).
To find the relative maximum and relative minimum of the graph of a function, you need to find the critical points and determine if they correspond to maximum or minimum points.
To find the critical points, we need to find the values of x where the derivative of the function is zero or undefined.
First, let's find the derivative of the function f(x) = x^3 + 2x^2 - 2x.
f'(x) = 3x^2 + 4x - 2
Now, set f'(x) equal to zero and solve for x:
3x^2 + 4x - 2 = 0
This is a quadratic equation that can be solved using factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 3, b = 4, and c = -2. Substituting these values into the formula:
x = (-4 ± √(4^2 - 4 * 3 * -2)) / (2 * 3)
Simplifying further:
x = (-4 ± √(16 + 24)) / 6
x = (-4 ± √40) / 6
x = (-4 ± 2√10) / 6
x = (-2 ± √10) / 3
So we have two critical points: x = (-2 + √10) / 3 and x = (-2 - √10) / 3.
Now, we need to determine if these critical points correspond to a relative maximum or a relative minimum. For that, we can use the second derivative test.
Take the second derivative of the function:
f''(x) = 6x + 4
Now, substitute the critical points into the second derivative:
f''((-2 + √10) / 3) = 6((-2 + √10) / 3) + 4
f''((-2 + √10) / 3) = -2 + 2√10 + 4
f''((-2 + √10) / 3) = 2√10 + 2
Since the second derivative is positive at x = (-2 + √10) / 3, this indicates a relative minimum at that point.
Next, substitute the other critical point into the second derivative:
f''((-2 - √10) / 3) = 6((-2 - √10) / 3) + 4
f''((-2 - √10) / 3) = -2 - 2√10 + 4
f''((-2 - √10) / 3) = -2√10 + 2
Since the second derivative is negative at x = (-2 - √10) / 3, this indicates a relative maximum at that point.
Now, we can plug the critical points back into the original function to find their corresponding y-values:
f((-2 + √10) / 3) = ((-2 + √10) / 3)^3 + 2((-2 + √10) / 3)^2 - 2((-2 + √10) / 3)
f((-2 + √10) / 3) ≈ -1.65
f((-2 - √10) / 3) = ((-2 - √10) / 3)^3 + 2((-2 - √10) / 3)^2 - 2((-2 - √10) / 3)
f((-2 - √10) / 3) ≈ -0.42
Therefore, the relative maximum occurs at approximately (-2.68, -1.65), and the relative minimum occurs at approximately (0.39, -0.42).