Solving inverse function

Question

$$f : \Bbb R \rightarrow \Bbb R $$

$$f^{-1}(2x-7) = x-1, f(a-1) = 5$$

Determine $a$.

The inverse of the function $f^{-1}(2x-7)$ is written as

$$f\biggr (\dfrac{x+7}{2}\biggr ) = x-1$$

Now we have that

$$x+7 = 2(a-1)$$

Solving for x and we get

$$x = 2a-9$$

Plugging $x$ into the function

$$f(a-2) = 2a-9 = 5$$

Solving for $a$

$$2a-9 = 5 \implies 2a = 14 \implies a = 7$$

Answer 1

I think you are making a mistake when it comes to taking the inverse. If we say that $$ f:\mathbb{R}\rightarrow\mathbb{R} $$ $$ g(x) = 2x-7 $$ then $$ f^{-1}(2x-7) \equiv f^{-1}\circ g $$ where you are incorrectly saying that $$ f^{-1} \equiv g $$ now to actually find the values, if $$ f^{-1}(2x-7) = x-1 \implies f(x-1) = 2x-7 $$ in order to find the function, we just do a quick observation to see that $$ f(x) = 2x - 5 \therefore f(x-1)=2(x-2)-5 = 2x-2-5 = 2x-7 $$ Finally, to solve your problem we find that $$ f(a-1)=5 \implies 2(a-1)-5 = 5 \implies 2(a-1)=10 \implies a-1 = 5 $$ Therefore $a=6$

Answer 2

From the definition of $f^{-1}$, it's clear that $f(x-1) = 2x-7$. Substitute $x = a$ into $f(a-1) = 5$ to get $2a - 7 = 5.$ Therefore, $a = 6$.

Answer 3

You make it too complex. From the given,

$$f^{-1}(2a-7)=a-1\implies2a-7=f(a-1)=5.$$