Find the relative maximum and relative minimum of the graph of the function.
f(x)=x^3+5x^2-2x
To find the relative maximum and relative minimum of the graph of the function f(x) = x^3 + 5x^2 - 2x, we need to find the critical points and determine if they correspond to a minimum or maximum.
1. Find the derivative of f(x) with respect to x.
f'(x) = 3x^2 + 10x - 2
2. Set f'(x) = 0 and solve for x to find the critical points.
3x^2 + 10x - 2 = 0
Using the quadratic formula, we have:
x = (-b ± √(b^2 - 4ac)) / (2a)
x = (-10 ± √(10^2 - 4(3)(-2))) / (2(3))
x = (-10 ± √(100 + 24)) / 6
x = (-10 ± √124) / 6
x = (-10 ± 2√31) / 6
x = (-5 ± √31) / 3
The critical points are x = (-5 + √31) / 3 and x = (-5 - √31) / 3.
3. To determine if these are relative maximum or minimum points, we can examine the sign of the derivative around these points.
3a. For x = (-5 + √31) / 3:
- Plug in a value slightly less than (-5 + √31) / 3 into f'(x):
f'(-2) = 3(-2)^2 + 10(-2) - 2 = 12 - 20 - 2 = -10
- Plug in a value slightly greater than (-5 + √31) / 3 into f'(x):
f'(-1) = 3(-1)^2 + 10(-1) - 2 = 3 - 10 - 2 = -9
Since the values of f'(x) are negative on both sides of x = (-5 + √31) / 3, it indicates a relative maximum at this point.
3b. For x = (-5 - √31) / 3:
- Plug in a value slightly less than (-5 - √31) / 3 into f'(x):
f'(-3) = 3(-3)^2 + 10(-3) - 2 = 27 - 30 - 2 = -5
- Plug in a value slightly greater than (-5 - √31) / 3 into f'(x):
f'(-2) = 3(-2)^2 + 10(-2) - 2 = 12 - 20 - 2 = -10
Since the values of f'(x) are negative on both sides of x = (-5 - √31) / 3, it indicates a relative maximum at this point.
Therefore, the relative maximum occurs at x = (-5 + √31) / 3 and x = (-5 - √31) / 3. There are no relative minimum points for this function.
To find the relative maximum and relative minimum of the graph of the function f(x) = x^3 + 5x^2 - 2x, we need to find the critical points and determine their nature.
Step 1: Find the derivative of the function f'(x).
f'(x) = 3x^2 + 10x - 2
Step 2: Set f'(x) = 0 to find the critical points.
3x^2 + 10x - 2 = 0
Step 3: Solve the quadratic equation.
Using the quadratic formula, we have:
x = (-b ± √(b^2 - 4ac)) / 2a
x = (-10 ± √(10^2 - 4(3)(-2))) / (2)(3)
x = (-10 ± √(100 + 24)) / 6
x = (-10 ± √124) / 6
x ≈ -1.92 or x ≈ 0.26
Step 4: Determine the nature of the critical points.
To determine if a critical point is a relative maximum or relative minimum, we need to check the second derivative.
Step 4.1: Find the second derivative of f(x).
f''(x) = 6x + 10
Step 4.2: Substitute the critical points into the second derivative.
For x ≈ -1.92:
f''(-1.92) = 6(-1.92) + 10
f''(-1.92) ≈ -1.52
Since f''(-1.92) < 0, the point x ≈ -1.92 corresponds to a relative maximum.
For x ≈ 0.26:
f''(0.26) = 6(0.26) + 10
f''(0.26) ≈ 11.56
Since f''(0.26) > 0, the point x ≈ 0.26 corresponds to a relative minimum.
Step 5: Determine the y-values corresponding to the critical points.
For x ≈ -1.92:
f(-1.92) = (-1.92)^3 + 5(-1.92)^2 - 2(-1.92)
f(-1.92) ≈ -8.7
For x ≈ 0.26:
f(0.26) = (0.26)^3 + 5(0.26)^2 - 2(0.26)
f(0.26) ≈ 0.19
Step 6: State the relative maximum and relative minimum.
The relative maximum occurs at approximately (-1.92, -8.7), and the relative minimum occurs at approximately (0.26, 0.19).