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).
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)
1 answer