Question

If $2x$ is defined as inverse function of $x^2-5x+6, x\le \frac{5}{2}$

Then find

$1>20$

Answer 1

-2

Answer 2

We are given the function

$$f(x)=x^2-5x+6 \quad \text{with} \quad x\le \frac{5}{2}.$$

The inverse function $f^{-1}$ is defined on the range of $f$ for this restricted domain. To find $f^{-1}(20)$, solve for $x$ in:

$$x^2-5x+6=20.$$

Subtract 20 from both sides:

$$x^2-5x-14=0.$$

Factor or use the quadratic formula:

$$x=\frac{5\pm\sqrt{25+56}}{2}=\frac{5\pm\sqrt{81}}{2}=\frac{5\pm 9}{2}.$$

So, the solutions are:

$$x= \frac{5+9}{2}=7 \quad \text{and} \quad x=\frac{5-9}{2}=-2.$$

Since the original function has domain $x\le \frac{5}{2}$, we choose $x=-2$.

Thus,

$$f^{-1}(20)=-2.$$