Question

Find a cubic function f(x) = ax3 + bx2 + cx + d that has a local maximum value of 3 at x = −2 and a local minimum value of 0 at x = 1. f(x) =

Answer 1

Answer:

Step-by-step explanation:

f(-2)=3

f(1) =0

f'(-2)=0

f' (1) = 0 ..... continue

Answer 2

$f(x)=ax^3+bx^2+cx+d$

$\implies f'(x)=3ax^2+2bx+c$

Local extrema occur wherever $f'(x)=0$.

We want

• a local max of 3 when $x=-2$, which means

$f(-2)=-8a+4b-2c+d=3$

$f'(-2)=12a-4b=0\implies3a=b$

and

• a local min of 0 when $x=1$, which means

$f(1)=a+b+c+d=0$

$f'(1)=3a+2b+c=0$

Solve the system for $a,b,c,d$; you should get

$a=\dfrac19,b=\dfrac13,c=-1,d=\dfrac59$

so that the cubic we're looking for is

$\boxed{f(x)=\dfrac{x^3}9+\dfrac{x^2}3-x+\dfrac59}$