# (The function is cubed) How do you find a and b?

## $f \left(x\right) = 3 {\left(a x - \frac{b}{x}\right)}^{3}$ Given $f$$\left(\frac{3}{2}\right)$ = $3$ and $f '$$\left(\frac{3}{2}\right)$ = 30

Feb 4, 2018

a=2;b=3

#### Explanation:

Since
$f \left(\frac{3}{2}\right) = 3 {\left(\frac{3}{2} a - \frac{b}{\frac{3}{2}}\right)}^{3}$
we get:
$\cancel{3} {\left(\frac{3}{2} a - \frac{b}{\frac{3}{2}}\right)}^{3} = {\cancel{3}}^{1}$
that's

$\textcolor{red}{\frac{3}{2} a - \frac{2}{3} b = 1}$

Since
$f ' \left(x\right) = 3 \cdot 3 {\left(a x - \frac{b}{x}\right)}^{2} \cdot \left(a + \frac{b}{x} ^ 2\right)$
we get
$f ' \left(\frac{3}{2}\right) = 9 {\left(\frac{3}{2} a - \frac{b}{\frac{3}{2}}\right)}^{2} \cdot \left(a + \frac{b}{\frac{3}{2}} ^ 2\right) = 30$
that's
${\cancel{9}}^{3} {\left(\frac{3}{2} a - \frac{2}{3} b\right)}^{2} \cdot \left(a + \frac{4}{9} b\right) = {\cancel{30}}^{10}$

Let's substitute $\textcolor{red}{\frac{3}{2} a - \frac{2}{3} b = 1}$, then we get:

$3 \left(a + \frac{4}{9} b\right) = 10$
that's

$\textcolor{b l u e}{3 a + \frac{4}{3} b = 10}$

Since $\textcolor{red}{\frac{3}{2} a - \frac{2}{3} b = 1} \to 3 a = 2 + \frac{4}{3} b$
and we'll substitute it in $\textcolor{b l u e}{3 a + \frac{4}{3} b = 10}$
The we get:

$\frac{4}{3} b + \frac{4}{3} b + 2 = 10$
$\frac{8}{3} b - 8 = 0$
$b = 3$

Since $a = \frac{2}{3} + \frac{4}{9} b$, we get:
$b = 3 \to a = \frac{2}{3} + \frac{4}{3} = 2$

Feb 4, 2018

#### Explanation:

$f \left(x\right) = 3 {\left(a x - \frac{b}{x}\right)}^{3}$

We are given: $f \left(\frac{3}{2}\right) = 3$

Using the definition of $f$, we get
$f \left(\frac{3}{2}\right) = 3 {\left(\frac{3}{2} a - \frac{b}{\frac{3}{2}}\right)}^{3} = 3$

So $\frac{3}{2} a - \frac{2}{3} b = 1$.

Clear fractions by multiplying by $6$ to get

$\textcolor{red}{9 a - 4 b = 6}$

We are also given some information about the derivative of $f$, so
we find the derivative:

$f ' \left(x\right) = 9 {\left(a x - \frac{b}{x}\right)}^{2} \left(a + \frac{b}{x} ^ 2\right)$

Using $x = \frac{3}{2}$, gives us

$f ' \left(\frac{3}{2}\right) = 9 {\left(\frac{3}{2} a - \frac{2}{3} b\right)}^{2} \left(a + \frac{b}{\frac{3}{2}} ^ 2\right)$

$= 9 {\left(\frac{3}{2} a - \frac{2}{3} b\right)}^{2} \left(a + \frac{4}{9} b\right)$

$= {\left(\frac{3}{2} a - \frac{2}{3} b\right)}^{2} \left(9 a + 4 b\right)$

Recall from above that the first factor is $1$ and also we were given that $f ' \left(\frac{3}{2}\right) = 30$. So,

$f ' \left(\frac{3}{2}\right) = {\left(1\right)}^{2} \left(9 a + 4 b\right) = 30$ so

$\textcolor{b l u e}{9 a + 4 b = 30}$

Solving the system

$9 a + 4 b = 30$
$9 a - 4 b = 6$

We get

$18 a = 36$ so $a = 2$ and $8 b = 24$ so $b = 3$