1. ## Inverse Function Help

Hi, I have this question that I can't seem to work out.

$f(x) = 6x + 7x^{19}$
I need to find $a = f^{-1}(c)$ where $c = -13$

I can't seem to work out an inverse function.
I tried to say

$f(f^{-1}(-13)) = -13 = 6f^{-1}(-13) + 7(f^{-1}(-13))^{19}$
$-13 = 6a + 7a^{19}$

But how can I solve for a?

2. Originally Posted by alan4cult
Hi, I have this question that I can't seem to work out.

$f(x) = 6x + 7x^{19}$
I need to find $a = f^{-1}(c)$ where $c = -13$

I can't seem to work out an inverse function.
I tried to say

$f(f^{-1}(-13)) = -13 = 6f^{-1}(-13) + 7(f^{-1}(-13))^{19}$
$-13 = 6a + 7a^{19}$

But how can I solve for a?

You have the polynomial equation

$7a^{19}+6a+13=0$

From the rational roots theorem we now the only root are the factors of
13 $\{\pm 1,\pm 13 \}$ divided by the factors of 7 $\{\pm 1, \pm 7 \}$

So if we check this finite list we see that

$7(-1)^{19}+6(-1)+13=0$ this tells us that -1 is a solution to the equation.

We know this is the only solution because the derivative of f is alwasy posative so the function is 1-1.

Good luck. TES