Question

1. find the relative extrema of $f(x) = x^3 + 3x^2 - 9x + 5$

Explanation:

Step1: Find first derivative

$f'(x) = 3x^2 + 6x - 9$

Step2: Solve $f'(x)=0$

$3x^2 + 6x - 9 = 0$
Divide by 3: $x^2 + 2x - 3 = 0$
Factor: $(x+3)(x-1)=0$
Solutions: $x=-3, x=1$

Step3: Find second derivative

$f''(x) = 6x + 6$

Step4: Test $x=-3$ in $f''(x)$

$f''(-3) = 6(-3)+6 = -12 < 0$
So $x=-3$ is a relative maximum.
Calculate $f(-3)$:
$f(-3) = (-3)^3 + 3(-3)^2 -9(-3)+5 = -27 + 27 +27 +5 = 32$

Step5: Test $x=1$ in $f''(x)$

$f''(1) = 6(1)+6 = 12 > 0$
So $x=1$ is a relative minimum.
Calculate $f(1)$:
$f(1) = (1)^3 + 3(1)^2 -9(1)+5 = 1 + 3 -9 +5 = 0$

Answer:

Relative maximum at $x=-3$ with value $32$; relative minimum at $x=1$ with value $0$