# Composite function : If f(x)= 3x+1 and g(x) = x/(x^2+25), how to solve g(f(x))=0 ?

Aug 1, 2016

$g \left(f \left(x\right)\right) = 0 \text{ }$ for $\text{ } x = - \frac{1}{3}$

#### Explanation:

$f \left(x\right) = 3 x + 1$

$g \left(x\right) = \frac{x}{{x}^{2} + 25}$

The first thing to do here is figure out how the expression for

$\left(g \circ f\right) \left(x\right) = g \left(f \left(x\right)\right)$

To do that, use $f \left(x\right) = \textcolor{p u r p \le}{3 x + 1}$ as a value for $x$ in $g \left(x\right) = \frac{x}{{x}^{2} + 25}$. You will get

$g \left(f \left(x\right)\right) = \frac{\textcolor{p u r p \le}{3 x + 1}}{{\left(\textcolor{p u r p \le}{3 x + 1}\right)}^{2} + 25}$

$g \left(f \left(x\right)\right) = \frac{3 x + 1}{\left(9 {x}^{2} + 6 x + 1\right) + 25}$

$g \left(f \left(x\right)\right) = \frac{3 x + 1}{9 {x}^{2} + 6 x + 26}$

Now, you know that you must solve

$g \left(f \left(x\right)\right) = 0$

which means that you have

$\frac{3 x + 1}{9 {x}^{2} + 6 x + 26} = 0$

As you know, a fraction can be equal to zero only if its numerator is equal to zero. In your case, this implies

$3 x + 1 = 0 \implies \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{x = - \frac{1}{3}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

One last thing to do here -- make sure that $x = - \frac{1}{3}$ does not make the denominator equal to zero.

$9 \cdot {\left(- \frac{1}{3}\right)}^{2} + 6 \cdot \left(- \frac{1}{3}\right) + 26 = \textcolor{red}{\cancel{\textcolor{b l a c k}{9}}} \cdot \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{9}}}} - 2 + 26$

$= 1 - 2 + 26 \ne 0$

Therefore, you can say that $g \left(f \left(x\right)\right) = 0$ for $x = - \frac{1}{3}$.

As an alternative to plugging in $x = - \frac{1}{3}$ to check if the denominator is equal to zero, notice that this quadratic equation

$9 {x}^{2} + 6 x + 26 = 0$

will never actually be equal to zero because its discriminant

$\Delta = {b}^{2} - 4 \cdot a \cdot c$

$\Delta = {6}^{2} - 4 \cdot 9 \cdot 26$

is negative, $\Delta < 0$. This implies that this quadratic equation doesn't have real numbers as its solutions, i.e. it has two complex roots.

You thus have,

$9 {x}^{2} + 6 x + 26 > 0 \text{ } \left(\forall\right) x \in \mathbb{R}$