# Question #c8903

Feb 9, 2017

$f \left(x\right) = - \frac{1}{2} {x}^{3} + \frac{9}{2} {x}^{2} - \frac{15}{2} x - \frac{5}{2}$

#### Explanation:

Write the general cubic function:

$f \left(x\right) = a {x}^{3} + b {x}^{2} + c x + d$

Its first and second derivative are:

$f ' \left(x\right) = 3 a {x}^{2} + 2 b x + c$

$f ' ' \left(x\right) = 6 a x + 2 b$

If $f \left(x\right)$ has an inflection point in $\left(3 , 2\right)$ this means that:

$f ' ' \left(3\right) = 0 \implies \textcolor{b l u e}{18 a + 2 b = 0}$

and

$f \left(3\right) = 2 \implies \textcolor{b l u e}{27 a + 9 b + 3 c + d = 2}$

Then we know that the point $\left(5 , 10\right)$ is a maximum, so:

$f ' \left(5\right) = 0 \implies \textcolor{b l u e}{75 a + 10 b + c = 0}$

$f \left(5\right) = 10 \implies \textcolor{b l u e}{125 a + 25 b + 5 c + d = 10}$

We have now four linear equations in four unknowns: the solution of the system gives us the values of the coefficients of the cubic:

$\left\{\begin{matrix}18 a + 2 b = 0 \\ 27 a + 9 b + 3 c + d = 2 \\ 75 a + 10 b + c = 0 \\ 125 a + 25 b + 5 c + d = 10\end{matrix}\right.$

Subtract the second equation from the fourth:

$\left\{\begin{matrix}18 a + 2 b = 0 \\ 98 a + 16 b + 2 c = 8 \\ 75 a + 10 b + c = 0 \\ 125 a + 25 b + 5 c + d = 10\end{matrix}\right.$

Multiply the third equation by $2$:

$\left\{\begin{matrix}18 a + 2 b = 0 \\ 98 a + 16 b + 2 c = 8 \\ 150 a + 20 b + 2 c = 0 \\ 125 a + 25 b + 5 c + d = 10\end{matrix}\right.$

Subtract the second from the third:

$\left\{\begin{matrix}18 a + 2 b = 0 \\ 52 a + 4 b = - 8 \\ 150 a + 20 b + 2 c = 0 \\ 125 a + 25 b + 5 c + d = 10\end{matrix}\right.$

Multiply the first equation by $2$:

$\left\{\begin{matrix}36 a + 4 b = 0 \\ 52 a + 4 b = - 8 \\ 150 a + 20 b + 2 c = 0 \\ 125 a + 25 b + 5 c + d = 10\end{matrix}\right.$

Subtract the first from the second:

$\left\{\begin{matrix}16 a = - 8 \\ 52 a + 4 b = - 8 \\ 150 a + 20 b + 2 c = 0 \\ 125 a + 25 b + 5 c + d = 10\end{matrix}\right.$

and we have:

$\left\{\begin{matrix}a = - \frac{1}{2} \\ b = \frac{9}{2} \\ c = - \frac{15}{2} \\ d = - \frac{5}{2}\end{matrix}\right.$

So the function is:

$f \left(x\right) = - \frac{1}{2} {x}^{3} + \frac{9}{2} {x}^{2} - \frac{15}{2} x - \frac{5}{2}$

graph{-1/2x^3+9/2x^2-15/2x-5/2 [-29.24, 34.84, -14.1, 17.94]}