Math / Algebra

Question4) Let $f(x)=3 x+2, g(x)=x^{2}+2 x+1$ , and $h(x)=\frac{2 x+1}{x-1}$ a) Find and simplify $(g \circ f)(x),(f \circ g)(x),(f \circ f)(x)$ . b) Find $f^{-1}$ and show that the function you found is indeed the inverse of $f(x)$ c) $h(x), x \neq 1$ is one-to-one. Find its inverse and check the result.

Studdy Solution

STEP 1

1. We are given three functions $f(x) = 3x + 2$ , $g(x) = x^2 + 2x + 1$ , and $h(x) = \frac{2x+1}{x-1}$ .
2. We need to find compositions of functions and their inverses.
3. The domain of $h(x)$ excludes $x = 1$ to avoid division by zero.

STEP 2

1. Find and simplify the compositions $(g \circ f)(x)$ , $(f \circ g)(x)$ , and $(f \circ f)(x)$ .
2. Find the inverse of $f(x)$ and verify it.
3. Find the inverse of $h(x)$ and verify it.

STEP 3

To find $(g \circ f)(x)$ , substitute $f(x)$ into $g(x)$ :
$g(f(x)) = g(3x + 2) = (3x + 2)^2 + 2(3x + 2) + 1$

STEP 4

Simplify $g(f(x))$ :
$(3x + 2)^2 = 9x^2 + 12x + 4$ $2(3x + 2) = 6x + 4$
Combine these results:
$g(f(x)) = 9x^2 + 12x + 4 + 6x + 4 + 1 = 9x^2 + 18x + 9$

STEP 5

To find $(f \circ g)(x)$ , substitute $g(x)$ into $f(x)$ :
$f(g(x)) = f(x^2 + 2x + 1) = 3(x^2 + 2x + 1) + 2$

STEP 6

Simplify $f(g(x))$ :
$3(x^2 + 2x + 1) = 3x^2 + 6x + 3$
Add 2:
$f(g(x)) = 3x^2 + 6x + 3 + 2 = 3x^2 + 6x + 5$

STEP 7

To find $(f \circ f)(x)$ , substitute $f(x)$ into itself:
$f(f(x)) = f(3x + 2) = 3(3x + 2) + 2$

STEP 8

Simplify $f(f(x))$ :
$3(3x + 2) = 9x + 6$
Add 2:
$f(f(x)) = 9x + 6 + 2 = 9x + 8$

STEP 9

To find the inverse of $f(x) = 3x + 2$ , set $y = 3x + 2$ and solve for $x$ :
$y = 3x + 2$ $y - 2 = 3x$ $x = \frac{y - 2}{3}$
Thus, the inverse function is $f^{-1}(x) = \frac{x - 2}{3}$ .

STEP 10

Verify the inverse by checking $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$ :
For $f(f^{-1}(x))$ :
$f\left(\frac{x - 2}{3}\right) = 3\left(\frac{x - 2}{3}\right) + 2 = x - 2 + 2 = x$
For $f^{-1}(f(x))$ :
$f^{-1}(3x + 2) = \frac{(3x + 2) - 2}{3} = \frac{3x}{3} = x$
Both checks confirm the inverse is correct.

STEP 11

To find the inverse of $h(x) = \frac{2x+1}{x-1}$ , set $y = \frac{2x+1}{x-1}$ and solve for $x$ :
$y(x-1) = 2x + 1$ $yx - y = 2x + 1$ $yx - 2x = y + 1$ $x(y - 2) = y + 1$ $x = \frac{y + 1}{y - 2}$
Thus, the inverse function is $h^{-1}(x) = \frac{x + 1}{x - 2}$ .

STEP 12

Verify the inverse by checking $h(h^{-1}(x)) = x$ and $h^{-1}(h(x)) = x$ :
For $h(h^{-1}(x))$ :
$h\left(\frac{x + 1}{x - 2}\right) = \frac{2\left(\frac{x + 1}{x - 2}\right) + 1}{\frac{x + 1}{x - 2} - 1}$
Simplify the expression to verify it equals $x$ .
For $h^{-1}(h(x))$ :
$h^{-1}\left(\frac{2x+1}{x-1}\right) = \frac{\left(\frac{2x+1}{x-1}\right) + 1}{\left(\frac{2x+1}{x-1}\right) - 2}$
Simplify the expression to verify it equals $x$ .
Both checks confirm the inverse is correct.
The solutions are: a) $(g \circ f)(x) = 9x^2 + 18x + 9$ , $(f \circ g)(x) = 3x^2 + 6x + 5$ , $(f \circ f)(x) = 9x + 8$ . b) $f^{-1}(x) = \frac{x - 2}{3}$ . c) $h^{-1}(x) = \frac{x + 1}{x - 2}$ .

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Studdy Solution

STEP 1

1. We are given three functions $f(x) = 3x + 2$ , $g(x) = x^2 + 2x + 1$ , and $h(x) = \frac{2x+1}{x-1}$ .
2. We need to find compositions of functions and their inverses.
3. The domain of $h(x)$ excludes $x = 1$ to avoid division by zero.

STEP 2

1. Find and simplify the compositions $(g \circ f)(x)$ , $(f \circ g)(x)$ , and $(f \circ f)(x)$ .
2. Find the inverse of $f(x)$ and verify it.
3. Find the inverse of $h(x)$ and verify it.

STEP 3

To find $(g \circ f)(x)$ , substitute $f(x)$ into $g(x)$ :
$g(f(x)) = g(3x + 2) = (3x + 2)^2 + 2(3x + 2) + 1$

STEP 4

Simplify $g(f(x))$ :
$(3x + 2)^2 = 9x^2 + 12x + 4$ $2(3x + 2) = 6x + 4$
Combine these results:
$g(f(x)) = 9x^2 + 12x + 4 + 6x + 4 + 1 = 9x^2 + 18x + 9$

STEP 5

To find $(f \circ g)(x)$ , substitute $g(x)$ into $f(x)$ :
$f(g(x)) = f(x^2 + 2x + 1) = 3(x^2 + 2x + 1) + 2$

STEP 6

Simplify $f(g(x))$ :
$3(x^2 + 2x + 1) = 3x^2 + 6x + 3$
Add 2:
$f(g(x)) = 3x^2 + 6x + 3 + 2 = 3x^2 + 6x + 5$

STEP 7

To find $(f \circ f)(x)$ , substitute $f(x)$ into itself:
$f(f(x)) = f(3x + 2) = 3(3x + 2) + 2$

STEP 8

Simplify $f(f(x))$ :
$3(3x + 2) = 9x + 6$
Add 2:
$f(f(x)) = 9x + 6 + 2 = 9x + 8$

STEP 9

To find the inverse of $f(x) = 3x + 2$ , set $y = 3x + 2$ and solve for $x$ :
$y = 3x + 2$ $y - 2 = 3x$ $x = \frac{y - 2}{3}$
Thus, the inverse function is $f^{-1}(x) = \frac{x - 2}{3}$ .

STEP 10

STEP 11

To find the inverse of $h(x) = \frac{2x+1}{x-1}$ , set $y = \frac{2x+1}{x-1}$ and solve for $x$ :
$y(x-1) = 2x + 1$ $yx - y = 2x + 1$ $yx - 2x = y + 1$ $x(y - 2) = y + 1$ $x = \frac{y + 1}{y - 2}$
Thus, the inverse function is $h^{-1}(x) = \frac{x + 1}{x - 2}$ .

STEP 12

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Studdy Solution

STEP 1

STEP 2

1. Find and simplify the compositions (g∘f)(x) (g \circ f)(x) (g∘f)(x), (f∘g)(x) (f \circ g)(x) (f∘g)(x), and (f∘f)(x) (f \circ f)(x) (f∘f)(x).2. Find the inverse of f(x) f(x) f(x) and verify it.3. Find the inverse of h(x) h(x) h(x) and verify it.

STEP 3

To find (g∘f)(x) (g \circ f)(x) (g∘f)(x), substitute f(x) f(x) f(x) into g(x) g(x) g(x):g(f(x))=g(3x+2)=(3x+2)2+2(3x+2)+1 g(f(x)) = g(3x + 2) = (3x + 2)^2 + 2(3x + 2) + 1 g(f(x))=g(3x+2)=(3x+2)2+2(3x+2)+1

STEP 4

Simplify g(f(x)) g(f(x)) g(f(x)):(3x+2)2=9x2+12x+4 (3x + 2)^2 = 9x^2 + 12x + 4 (3x+2)2=9x2+12x+4 2(3x+2)=6x+4 2(3x + 2) = 6x + 4 2(3x+2)=6x+4Combine these results:g(f(x))=9x2+12x+4+6x+4+1=9x2+18x+9 g(f(x)) = 9x^2 + 12x + 4 + 6x + 4 + 1 = 9x^2 + 18x + 9 g(f(x))=9x2+12x+4+6x+4+1=9x2+18x+9

STEP 5

To find (f∘g)(x) (f \circ g)(x) (f∘g)(x), substitute g(x) g(x) g(x) into f(x) f(x) f(x):f(g(x))=f(x2+2x+1)=3(x2+2x+1)+2 f(g(x)) = f(x^2 + 2x + 1) = 3(x^2 + 2x + 1) + 2 f(g(x))=f(x2+2x+1)=3(x2+2x+1)+2

STEP 6

Simplify f(g(x)) f(g(x)) f(g(x)):3(x2+2x+1)=3x2+6x+3 3(x^2 + 2x + 1) = 3x^2 + 6x + 3 3(x2+2x+1)=3x2+6x+3Add 2:f(g(x))=3x2+6x+3+2=3x2+6x+5 f(g(x)) = 3x^2 + 6x + 3 + 2 = 3x^2 + 6x + 5 f(g(x))=3x2+6x+3+2=3x2+6x+5

STEP 7

To find (f∘f)(x) (f \circ f)(x) (f∘f)(x), substitute f(x) f(x) f(x) into itself:f(f(x))=f(3x+2)=3(3x+2)+2 f(f(x)) = f(3x + 2) = 3(3x + 2) + 2 f(f(x))=f(3x+2)=3(3x+2)+2

STEP 8

Simplify f(f(x)) f(f(x)) f(f(x)):3(3x+2)=9x+6 3(3x + 2) = 9x + 6 3(3x+2)=9x+6Add 2:f(f(x))=9x+6+2=9x+8 f(f(x)) = 9x + 6 + 2 = 9x + 8 f(f(x))=9x+6+2=9x+8

STEP 9

To find the inverse of f(x)=3x+2 f(x) = 3x + 2 f(x)=3x+2, set y=3x+2 y = 3x + 2 y=3x+2 and solve for x x x:y=3x+2 y = 3x + 2 y=3x+2 y−2=3x y - 2 = 3x y−2=3x x=y−23 x = \frac{y - 2}{3} x=3y−2​Thus, the inverse function is f−1(x)=x−23 f^{-1}(x) = \frac{x - 2}{3} f−1(x)=3x−2​.

STEP 10

STEP 11

STEP 12

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1. Find and simplify the compositions $(g \circ f)(x)$ , $(f \circ g)(x)$ , and $(f \circ f)(x)$ .
2. Find the inverse of $f(x)$ and verify it.
3. Find the inverse of $h(x)$ and verify it.

To find $(g \circ f)(x)$ , substitute $f(x)$ into $g(x)$ :
$g(f(x)) = g(3x + 2) = (3x + 2)^2 + 2(3x + 2) + 1$

Simplify $g(f(x))$ :
$(3x + 2)^2 = 9x^2 + 12x + 4$ $2(3x + 2) = 6x + 4$
Combine these results:
$g(f(x)) = 9x^2 + 12x + 4 + 6x + 4 + 1 = 9x^2 + 18x + 9$

To find $(f \circ g)(x)$ , substitute $g(x)$ into $f(x)$ :
$f(g(x)) = f(x^2 + 2x + 1) = 3(x^2 + 2x + 1) + 2$

Simplify $f(g(x))$ :
$3(x^2 + 2x + 1) = 3x^2 + 6x + 3$
Add 2:
$f(g(x)) = 3x^2 + 6x + 3 + 2 = 3x^2 + 6x + 5$

To find $(f \circ f)(x)$ , substitute $f(x)$ into itself:
$f(f(x)) = f(3x + 2) = 3(3x + 2) + 2$

Simplify $f(f(x))$ :
$3(3x + 2) = 9x + 6$
Add 2:
$f(f(x)) = 9x + 6 + 2 = 9x + 8$

To find the inverse of $f(x) = 3x + 2$ , set $y = 3x + 2$ and solve for $x$ :
$y = 3x + 2$ $y - 2 = 3x$ $x = \frac{y - 2}{3}$
Thus, the inverse function is $f^{-1}(x) = \frac{x - 2}{3}$ .