QuestionGiven the following functions: - $f(x)=\frac{4 x-3}{x-2}$ - $g(x)=2 x-1$
Determine the rule of $g \circ f(x+h)$ .

Studdy Solution

STEP 1

1. The composition of functions $g \circ f$ means applying $f$ first, then applying $g$ to the result.
2. We need to find $g(f(x+h))$ , which involves substituting $x+h$ into $f$ and then applying $g$ .

STEP 2

1. Substitute $x+h$ into the function $f(x)$ .
2. Simplify the expression for $f(x+h)$ .
3. Apply the function $g$ to the result of $f(x+h)$ .
4. Simplify the expression for $g(f(x+h))$ .

STEP 3

Substitute $x+h$ into the function $f(x)$ :
$f(x+h) = \frac{4(x+h) - 3}{(x+h) - 2}$

STEP 4

Simplify the expression for $f(x+h)$ :
$f(x+h) = \frac{4x + 4h - 3}{x + h - 2}$

STEP 5

Apply the function $g$ to the result of $f(x+h)$ . We need to substitute $f(x+h)$ into $g(x)$ :
$g(f(x+h)) = g\left(\frac{4x + 4h - 3}{x + h - 2}\right)$
Since $g(x) = 2x - 1$ , substitute:
$g\left(\frac{4x + 4h - 3}{x + h - 2}\right) = 2\left(\frac{4x + 4h - 3}{x + h - 2}\right) - 1$

STEP 6

Simplify the expression for $g(f(x+h))$ :
$g(f(x+h)) = \frac{2(4x + 4h - 3)}{x + h - 2} - 1$
$g(f(x+h)) = \frac{8x + 8h - 6}{x + h - 2} - 1$
To combine into a single fraction:
$g(f(x+h)) = \frac{8x + 8h - 6 - (x + h - 2)}{x + h - 2}$
$g(f(x+h)) = \frac{8x + 8h - 6 - x - h + 2}{x + h - 2}$
$g(f(x+h)) = \frac{7x + 7h - 4}{x + h - 2}$
The rule of $g \circ f(x+h)$ is $\boxed{\frac{7x + 7h - 4}{x + h - 2}}$ .

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