Find the relative maximum and relative minimum of the graph of the function.
f(x)=x^3+2x^2-5x
Determine where the relative maximum occurs and round each coordinate to the nearest hundredth. Choose the correct answer below.
(0.29,8.94)
(0.79,-2.21)
(-1.16,-0.86)
(-2.12,10.06)
To find the relative maximum and minimum of the function, we first need to find the critical points. The critical points occur when the derivative of the function is equal to zero or is undefined. Taking the derivative of the function, we get:
f'(x) = 3x^2 + 4x - 5
Setting f'(x) = 0, we can solve for x:
3x^2 + 4x - 5 = 0
Using the quadratic formula, we get:
x = (-4 ± √(4^2 - 4(3)(-5))) / (2(3))
x = (-4 ± √(16 + 60)) / 6
x = (-4 ± √76) / 6
x = (-4 ± 2√19) / 6
Simplifying, we have:
x = (-2 ± √19) / 3
So, we have two critical points: (-2 + √19) / 3 and (-2 - √19) / 3.
To determine whether these critical points are relative maxima or minima, we can use the second derivative test. Taking the derivative of f'(x), we get:
f''(x) = 6x + 4
To evaluate f''(x) at the critical points, we have:
f''((-2 + √19) / 3) = 6((-2 + √19) / 3) + 4
f''((-2 + √19) / 3) = -2 + 2√19 + 4
f''((-2 + √19) / 3) = 2 + 2√19
Since f''((-2 + √19) / 3) is positive, the critical point (-2 + √19) / 3 corresponds to a relative minimum.
Similarly, we have:
f''((-2 - √19) / 3) = 2 - 2√19
Since f''((-2 - √19) / 3) is negative, the critical point (-2 - √19) / 3 corresponds to a relative maximum.
Now, we can substitute the x-values of the critical points back into the original function to find the corresponding y-values.
f((-2 + √19) / 3) = ((-2 + √19) / 3)^3 + 2((-2 + √19) / 3)^2 - 5((-2 + √19) / 3)
f((-2 + √19) / 3) ≈ -0.86
So the relative maximum occurs at (-1.16, -0.86) (rounded to the nearest hundredth).
f((-2 - √19) / 3) = ((-2 - √19) / 3)^3 + 2((-2 - √19) / 3)^2 - 5((-2 - √19) / 3)
f((-2 - √19) / 3) ≈ 10.06
So the relative minimum occurs at (-2.12, 10.06) (rounded to the nearest hundredth).
Therefore, the correct answer is: (-2.12, 10.06).
To find the relative maximum and relative minimum of the given function f(x) = x^3 + 2x^2 - 5x, we need to find the critical points and then evaluate the function at those points.
Step 1: Find the derivative of the function f'(x) = 3x^2 + 4x - 5.
Step 2: Set the derivative equal to zero and solve for x to find the critical points:
3x^2 + 4x - 5 = 0
Step 3: Solve the quadratic equation using factoring, completing the square, or the quadratic formula.
The solutions to the quadratic equation are x = -1.16 and x = 0.79.
Step 4: Evaluate the function f(x) at the critical points to determine the nature of the stationary points.
For x = -1.16:
f(-1.16) = (-1.16)^3 + 2(-1.16)^2 - 5(-1.16) = -0.86
So, the relative minimum occurs at approximately (-1.16, -0.86).
For x = 0.79:
f(0.79) = (0.79)^3 + 2(0.79)^2 - 5(0.79) = 8.94
So, the relative maximum occurs at approximately (0.79, 8.94).
Therefore, the correct answer is (0.79, 8.94).