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

##### 1 Answer

#### Explanation:

Your two functions are

#f(x) = 3x + 1#

#g(x) = x/(x^2 + 25)#

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

#(g @ f)(x) = g(f(x))#

To do that, use

#g(f(x)) = color(purple)(3x+1)/( (color(purple)(3x+1))^2 + 25)#

#g(f(x)) = (3x + 1)/( (9x^2 + 6x + 1) + 25)#

#g(f(x)) = (3x+1)/(9x^2 + 6x + 26)#

Now, you know that you must solve

#g(f(x)) = 0#

which means that you have

#(3x+1)/(9x^2 + 6x + 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

#3x + 1 = 0 implies color(green)(|bar(ul(color(white)(a/a)color(black)(x = -1/3)color(white)(a/a)|)))#

One last thing to do here -- make sure that **does not** make the denominator equal to zero.

#9 * (-1/3)^2 + 6 * (-1/3) + 26 = color(red)(cancel(color(black)(9))) * 1/color(red)(cancel(color(black)(9))) - 2 + 26#

#=1 - 2 + 26 !=0#

Therefore, you can say that

As an alternative to plugging in

#9x^2 + 6x + 26 = 0#

will **never** actually be equal to zero because its discriminant

#Delta = b^2 - 4 * a * c#

#Delta = 6^2 - 4 * 9 * 26#

is **negative**, *real numbers* as its solutions, i.e. it has two *complex roots*.

You thus have,

#9x^2 + 6x + 26 > 0 " "(AA) x in RR#