Question

what is $f^{-1}(x)$?
$f(x)=\frac{5x + 2}{7}$
$f^{-1}(x)=\frac{?x+}{}$

Explanation:

Step1: Let \( y = f(x) \)

\( y=\frac{5x + 2}{7} \)

Step2: Solve for \( x \) in terms of \( y \)

Multiply both sides by \( 7 \): \( 7y=5x + 2 \)
Subtract \( 2 \): \( 5x=7y - 2 \)
Divide by \( 5 \): \( x=\frac{7y - 2}{5}=\frac{7}{5}y-\frac{2}{5} \)

Step3: Swap \( x \) and \( y \)

\( f^{-1}(x)=\frac{7x- 2}{5} \)

Answer:

The numerator of the first - degree term is \( 7 \), the constant term in the numerator is \(-2\), and the denominator is \(5\). So \( f^{-1}(x)=\frac{7x-2}{5}\), which means the values are \(7\), \(- 2\), \(5\) (from left - to - right in the fraction structure).