Question

Find a, b, c, and d such that the cubic function f(x) = ax3 + bx? + cx + d satisfies the given conditions.
Relative maximum: (2,9)
Relative minimum: (4,3)
Inflection point: (3,6)
a =
b =
C=
d =

Answer 1

Answer:

$\displaystyle f(x)=\frac{3}{2}x^3-\frac{27}{2}x^2+36x-21$

Where:

$\displaystyle a=\frac{3}{2}, \, b=-\frac{27}{2}, \, c=36, \text{and } d=-21$

Step-by-step explanation:

We are given a cubic function:

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

And we want to find a, b, c and d such that the  function has a relative maximum at (2, 9); a relative mininum at (4, 3); and an inflection point at (3, 6).

Since the function has a relative maximum at (2, 9), this means that:

$f(2)=9=a(2)^3+b(2)^2+c(2)+d$

Simplify:

$8a+4b+2c+d=9$

Likewise, since it has a relative minimum at (4, 3):

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

Simplify:

$64a+16b+4c+d=3$

We can subtract the first equation from the second. So:

$(64a+16b+4c+d)-(8a+4b+2c+d)=(3)-(9)$

Simplify:

$56a+12b+2c=-6$

Divide both sides by two. Hence:

$28a+6b+c=-3$

Relative minima occurs only at the critical points of a function. That is, it occurs whenever the first derivative equals zero.

Find the first derivative. We can treat a, b, c and d as constant. Hence:

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

Since it has a minima at (2, 9), it means that:

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

Thus:

$12a+4b+c=0$

(We will only need one of the two points to complete the problem.)

Inflection points occurs whenever the second derivative of a function equals zero. Find the second derivative:

$f''(x)=6ax+2b$

Since there is a inflection point at (3, 6):

$18a+2b=0\Rightarrow 9a+b=0$

Solve for b:

$b=-9a$

Substitute this into the above equation:

$12a+4(-9a)+c=0$

Solve for c:

$c=24a$

Substitute b and c into the previously acquired equation:

$28a+6(-9a)+(24a)=-3$

Solve for a:

$\displaystyle -2a=-3\Rightarrow a=\frac{3}{2}$

Solve for b and c:

$\displaystyle b=-9\left(\frac{3}{2}\right)=-\frac{27}{2}\text{ and } c=24\left(\frac{3}{2}\right)=36$

Using either the very first or second equation, solve for d:

$\displaystyle 8\left(\frac{3}{2}\right)+4\left(-\frac{27}{2}\right)+2(36)+d=9$

Hence:

$d=-21$

Hence, our function is:

$\displaystyle f(x)=\frac{3}{2}x^3-\frac{27}{2}x^2+36x-21$