Find expert answers and community support for all your questions on IDNLearn.com. Get accurate and comprehensive answers from our network of experienced professionals.
Answer:
[tex]\displaystyle f(x)=\frac{3}{2}x^3-\frac{27}{2}x^2+36x-21[/tex]
Where:
[tex]\displaystyle a=\frac{3}{2}, \, b=-\frac{27}{2}, \, c=36, \text{and } d=-21[/tex]
Step-by-step explanation:
We are given a cubic function:
[tex]f(x)=ax^3+bx^2+cx+d[/tex]
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:
[tex]f(2)=9=a(2)^3+b(2)^2+c(2)+d[/tex]
Simplify:
[tex]8a+4b+2c+d=9[/tex]
Likewise, since it has a relative minimum at (4, 3):
[tex]f(4)=3=a(4)^3+b(4)^2+c(4)+d[/tex]
Simplify:
[tex]64a+16b+4c+d=3[/tex]
We can subtract the first equation from the second. So:
[tex](64a+16b+4c+d)-(8a+4b+2c+d)=(3)-(9)[/tex]
Simplify:
[tex]56a+12b+2c=-6[/tex]
Divide both sides by two. Hence:
[tex]28a+6b+c=-3[/tex]
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:
[tex]f'(x)=3ax^2+2bx+c[/tex]
Since it has a minima at (2, 9), it means that:
[tex]f'(2)=3a(2)^2+2b(2)+c=0[/tex]
Thus:
[tex]12a+4b+c=0[/tex]
(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:
[tex]f''(x)=6ax+2b[/tex]
Since there is a inflection point at (3, 6):
[tex]18a+2b=0\Rightarrow 9a+b=0[/tex]
Solve for b:
[tex]b=-9a[/tex]
Substitute this into the above equation:
[tex]12a+4(-9a)+c=0[/tex]
Solve for c:
[tex]c=24a[/tex]
Substitute b and c into the previously acquired equation:
[tex]28a+6(-9a)+(24a)=-3[/tex]
Solve for a:
[tex]\displaystyle -2a=-3\Rightarrow a=\frac{3}{2}[/tex]
Solve for b and c:
[tex]\displaystyle b=-9\left(\frac{3}{2}\right)=-\frac{27}{2}\text{ and } c=24\left(\frac{3}{2}\right)=36[/tex]
Using either the very first or second equation, solve for d:
[tex]\displaystyle 8\left(\frac{3}{2}\right)+4\left(-\frac{27}{2}\right)+2(36)+d=9[/tex]
Hence:
[tex]d=-21[/tex]
Hence, our function is:
[tex]\displaystyle f(x)=\frac{3}{2}x^3-\frac{27}{2}x^2+36x-21[/tex]