To find the relative maximum and minimum of a function, we need to find the critical points first. The critical points occur when the derivative of the function equals 0 or does not exist.
First, let's find the derivative of the function f(x) = x^3 + 2x^2 - 5x.
f'(x) = 3x^2 + 4x - 5
Next, let's set the derivative equal to 0 and solve for x:
3x^2 + 4x - 5 = 0
We can factor the quadratic equation:
(3x - 1)(x + 5) = 0
Setting each factor equal to 0, we find the critical points:
3x - 1 = 0 --> x = 1/3
x + 5 = 0 --> x = -5
Now, let's plug these critical points back into the original function to find the corresponding y-coordinates.
f(1/3) = (1/3)^3 + 2(1/3)^2 - 5(1/3) = 1/27 + 2/9 - 5/3 = -2.21
f(-5) = (-5)^3 + 2(-5)^2 - 5(-5) = -125 + 50 + 25 = -50
The relative maximum occurs at (0.79, -2.21) and the relative minimum occurs at (-2.12, -50). Neither of these options are listed as choices, so there may be an error in the given answer choices.
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)
Determine where the relative minimum occurs and round each coordinate to the nearest hundredth. Choose the correct answer below.
(-2.12,10.06)
(0.79,-2.21)
(-1.16,-0.86)
(0.29,8.94)
1 answer