Answer:
the rel. max. is at (-4, 80) and the rel. min at (2, -28) (second answer choice)
Step-by-step explanation:
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If f(x) = x³ + 3x² - 24x, the derivative f '(x) is 3x² + 6x - 24 = f '(x)
We set this = to 0, divide all terms by 3 to simplify it, and solve the resulting equation for x: 0 = 3(x² + 2x - 8) = 3(x + 4)(x - 2) = 0
Then the derivative is zero at x = -4 and at x = 2.
This shows the location of any relative min and max.
Find the second derivative: f "(x) = 3(2x + 2)
and check the sign of this second derivative at the critical values 2 and -4:
f "(2) is + and so we conclude we have a rel. min. at x = 2; and:
f "(-4) is - and ... rel max at x = -4.
Finally, evaluate f(x) at each 2 and -4 to find the y-values of the relative max and rel min:
f(-4) = 80. We now have enough info to conclude that the rel. max. is at (-4, 80) and the rel. min at (2, -28).
