Anyone please solve and explain part b

Answer:

f(2) = -6

f inverse of 1/2 is 17/2

Step-by-step explanation:

We are given a function [tex]\displaystyle \large{f(x) = \frac{6}{2x-5}}[/tex]

To solve for part (a), we have to substitute x = 2 in f(x).

[tex]\displaystyle \large{f(2)=\frac{6}{2(2)-5} \\\displaystyle \large{f(2)=\frac{6}{4-5}=\frac{6}{-1}=-6[/tex]

Therefore, f(2) is -6.

Now we solve for part (b), we see the notation which is a minor different similar to f(x). We see that there is exponent of -1 between f and 1/2. The part (b) indicates that the function is an inverse of f(x).

[tex]\displaystyle \large{y=f^{-1}(x) \longrightarrow x=f(y)}[/tex]

To solve for part (b), first we solve for x in the function. Let f(x) = y.

[tex]\displaystyle \large{y=\frac{6}{2x-5}}[/tex]

Multiply both sides by 2x-5.

[tex]\displaystyle \large{y(2x-5)=\frac{6}{2x-5}(2x-5)}\\\displaystyle \large{y(2x-5)=6}[/tex]

Divide both sides by y-term.

[tex]\displaystyle \large{\frac{y(2x-5)}{y}=\frac{6}{y}}\\\displaystyle \large{2x-5=\frac{6}{y}}\\\displaystyle \large{2x=\frac{6}{y}+5}\\\displaystyle \large{x=\frac{6}{2y}+\frac{5}{2}}\\\displaystyle \large{x=\frac{3}{y}+\frac{5}{2}}[/tex]

Then swap x and y which we receive [tex]\displaystyle \large{y=\frac{3}{x}+\frac{5}{2}}[/tex]

Therefore, [tex]\displaystyle \large{f(x)=\frac{6}{2x-5} \longrightarrow f^{-1}(x)=\frac{3}{x}+\frac{5}{2}}[/tex]

Thus, [tex]\displaystyle \large{f^{-1}(\frac{1}{2})=\frac{3}{\frac{1}{2}}+\frac{5}{2}}\\\displaystyle \large{f^{-1}(\frac{1}{2})=6+\frac{5}{2}}\\\displaystyle \large{f^{-1}(\frac{1}{2})=\frac{17}{2}}[/tex]

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