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

The graph of $f$ passes through the points (-2, 3) and (1, 0), so

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

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

Since these are the sites of local extrema, we know that $f'(-2)$ and $f'(1)$ are either zero or undefined. $f$ is a polynomial, so it's continuous and differentiable everywhere, so only the zero-case is relevant.

We have derivative

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

and so

$f'(-2) = 0 \implies 12a - 4b + c = 0$

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

Solve for $a,b,c,d$.

• In the first two equations, we can eliminate $d$.

$(-8a+4b-2c+d) - (a+b+c+d) = 3-0 \implies -9a + 3b - 3c = 3$

• Now eliminate $c$ by combining any two equations in $a,b,c$.

$(12a-4b+c) - (3a+2b+c) = 0-0 \implies 9a-6b = 0 \implies 3a-2b=0$

$3(3a+2b+c) + (-9a+3b-3c) = 3(0)+3 \implies 9b = 3$

Then we have

$9b = 3 \implies b = \dfrac13$

$3a-2b=0 \implies 3a = \dfrac23 \implies a=\dfrac29$

$3a+2b+c=0 \implies c = -\dfrac43$

$a+b+c+d=0 \implies d = \dfrac79$

and so the cubic function is

$\boxed{f(x) = \dfrac29 x^3 + \dfrac13 x^2 - \dfrac43 x + \dfrac79}$