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