## Calculus: Early Transcendentals 8th Edition

$f(x) = \frac{2}{9}x^3+\frac{1}{3}x^2-\frac{4}{3}x+\frac{7}{9}$
$f(x) = ax^3+bx^2+cx+d$ $f'(x) = 3ax^2+2bx+c$ The following is given in the question: $f'(-2) = f'(1) = 0$ $f(-2) = 3$ $f(1) = 0$ Then: 1. $f'(-2) = 3a(-2)^2+2b(-2)+c = 12a-4b+c = 0$ 2. $f'(1) = 3a(1)^2+2b(1)+c = 3a+2b+c = 0$ 3. $f(-2) = a(-2)^3+b(-2)^2+c(-2)+d = -8a+4b-2c+d = 3$ 4. $f(1) = a(1)^3+b(1)^2+c(1)+d = a+b+c+d = 0$ We can add Equation (1) and $2\times$ Equation (2): $18a+3c = 0$ $c = -6a$ We can subtract Equation (4) from Equation (3): $-9a+3b-3c = 3$ $-9a+3b-3(-6a) = 3$ $9a+3b = 3$ $3a+b=1$ $b = 1-3a$ We can add Equation (1) and Equation (3): $4a-c+d = 3$ $4a-(-6a)+d = 3$ $10a+d = 3$ $d = 3-10a$ We can use Equation (1) to find the value of $a$: $12a-4b+c = 0$ $12a-4(1-3a)+(-6a) = 0$ $18a-4 = 0$ $a = \frac{2}{9}$ We can find the value of $b$: $b = 1-3a$ $b = 1-3(\frac{2}{9})$ $b = \frac{1}{3}$ We can find the value of $c$: $c = -6a$ $c = -6(\frac{2}{9})$ $c = -\frac{4}{3}$ We can find the value of $d$: $d = 3-10a$ $d = 3-10(\frac{2}{9})$ $d = \frac{7}{9}$ Therefore: $f(x) = \frac{2}{9}x^3+\frac{1}{3}x^2-\frac{4}{3}x+\frac{7}{9}$