Question

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

Answer 1

`g(f(x)) = 0" "` for `" "x = -1/3`

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 `f(x) = color(purple)(3x+1)` as a value for `x` in `g(x) = x/(x^2 + 25)`. You will get

`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 `x=-1/3` 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 `g(f(x)) = 0` for `x = -1/3`.

As an alternative to plugging in `x=-1/3` to check if the denominator is equal to zero, notice that this quadratic equation

`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, `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,

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