Math / Algebra

Question10.1 Given $f(x)=3-x^{2}$ (with domain $\left.(-\infty, \infty)\right)$ , $g(x)=2-x($ with domain $(-\infty, \infty))$ , $h(x)=\frac{1}{x}$ (with domain $\left.(0, \infty)\right)$ , find the following compositions (a) $f \circ g$ (b) $g \circ f$ (c) $f \circ h$ (d) $g \circ h$ (+) hof; What is the domain this function? $(+)$ hog; What is the domain this function? 10.2 Determine the inverses of the following functions (a) $f(x)=4-5 x$ , (with domain $(-\infty, \infty)$ ) (b) $h(x)=x^{2}-3 x+2$ , (with domain $\left.\left.(2, \infty)\right)\right]$ $(+) f(x)=\frac{2 x+1}{3-x}$ , (also find the domain and range of $\ell$ and of $f^{-1}$ )

Studdy Solution

STEP 1

1. The compositions of functions involve substituting one function into another.
2. The domain of a composition is determined by the domain of the inner function and the resulting domain after substitution.
3. Inverses of functions are found by solving the equation $y = f(x)$ for $x$ .

STEP 2

1. Find the compositions of functions. - Evaluate $f \circ g$ . - Evaluate $g \circ f$ . - Evaluate $f \circ h$ . - Evaluate $g \circ h$ . - Evaluate $h \circ f$ . - Evaluate $h \circ g$ .
2. Determine the inverses of functions. - Find the inverse of $f(x) = 4 - 5x$ . - Find the inverse of $h(x) = x^2 - 3x + 2$ . - Find the inverse of $f(x) = \frac{2x+1}{3-x}$ and determine the domain and range.

STEP 3

Evaluate $f \circ g$ .
$f \circ g(x) = f(g(x)) = f(2-x) = 3 - (2-x)^2$
Simplify the expression:
$= 3 - (4 - 4x + x^2) = 3 - 4 + 4x - x^2 = -1 + 4x - x^2$
The domain of $f \circ g$ is all real numbers, $(-\infty, \infty)$ , since $g(x)$ is defined for all $x$ .

STEP 4

Evaluate $g \circ f$ .
$g \circ f(x) = g(f(x)) = g(3-x^2) = 2 - (3-x^2)$
Simplify the expression:
$= 2 - 3 + x^2 = x^2 - 1$
The domain of $g \circ f$ is all real numbers, $(-\infty, \infty)$ , since $f(x)$ is defined for all $x$ .

STEP 5

Evaluate $f \circ h$ .
$f \circ h(x) = f(h(x)) = f\left(\frac{1}{x}\right) = 3 - \left(\frac{1}{x}\right)^2$
The domain of $f \circ h$ is $(0, \infty)$ , since $h(x)$ is defined for $x > 0$ .

STEP 6

Evaluate $g \circ h$ .
$g \circ h(x) = g(h(x)) = g\left(\frac{1}{x}\right) = 2 - \frac{1}{x}$
The domain of $g \circ h$ is $(0, \infty)$ , since $h(x)$ is defined for $x > 0$ .

STEP 7

Evaluate $h \circ f$ .
$h \circ f(x) = h(f(x)) = h(3-x^2) = \frac{1}{3-x^2}$
The domain of $h \circ f$ is determined by $3-x^2 > 0$ , which simplifies to $-\sqrt{3} < x < \sqrt{3}$ .

STEP 8

Evaluate $h \circ g$ .
$h \circ g(x) = h(g(x)) = h(2-x) = \frac{1}{2-x}$
The domain of $h \circ g$ is $x \neq 2$ , or $(-\infty, 2) \cup (2, \infty)$ .

STEP 9

Find the inverse of $f(x) = 4 - 5x$ .
Set $y = 4 - 5x$ and solve for $x$ :
$y = 4 - 5x$ $5x = 4 - y$ $x = \frac{4 - y}{5}$
Thus, the inverse is $f^{-1}(x) = \frac{4 - x}{5}$ .

STEP 10

Find the inverse of $h(x) = x^2 - 3x + 2$ .
Since the domain is $(2, \infty)$ , we solve $y = x^2 - 3x + 2$ for $x$ using the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
For $y = x^2 - 3x + 2$ , $a = 1$ , $b = -3$ , $c = 2 - y$ .
$x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (2-y)}}{2 \cdot 1}$ $x = \frac{3 \pm \sqrt{9 - 8 + 4y}}{2}$ $x = \frac{3 \pm \sqrt{1 + 4y}}{2}$
Since the domain is $(2, \infty)$ , we take the positive root:
$x = \frac{3 + \sqrt{1 + 4y}}{2}$
Thus, the inverse is $h^{-1}(x) = \frac{3 + \sqrt{1 + 4x}}{2}$ .

STEP 11

Find the inverse of $f(x) = \frac{2x+1}{3-x}$ .
Set $y = \frac{2x+1}{3-x}$ and solve for $x$ :
$y(3-x) = 2x + 1$ $3y - yx = 2x + 1$ $3y - 1 = 2x + yx$ $3y - 1 = x(2 + y)$ $x = \frac{3y - 1}{2 + y}$
Thus, the inverse is $f^{-1}(x) = \frac{3x - 1}{2 + x}$ .
The domain of $f$ is $x \neq 3$ , or $(-\infty, 3) \cup (3, \infty)$ .
The range of $f$ is all real numbers except the value that makes the denominator zero in the inverse, which is $y \neq -2$ .
The domain of $f^{-1}$ is $x \neq -2$ , or $(-\infty, -2) \cup (-2, \infty)$ .

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Studdy Solution

STEP 1

1. The compositions of functions involve substituting one function into another.
2. The domain of a composition is determined by the domain of the inner function and the resulting domain after substitution.
3. Inverses of functions are found by solving the equation $y = f(x)$ for $x$ .

STEP 2

STEP 3

Evaluate $f \circ g$ .
$f \circ g(x) = f(g(x)) = f(2-x) = 3 - (2-x)^2$
Simplify the expression:
$= 3 - (4 - 4x + x^2) = 3 - 4 + 4x - x^2 = -1 + 4x - x^2$
The domain of $f \circ g$ is all real numbers, $(-\infty, \infty)$ , since $g(x)$ is defined for all $x$ .

STEP 4

Evaluate $g \circ f$ .
$g \circ f(x) = g(f(x)) = g(3-x^2) = 2 - (3-x^2)$
Simplify the expression:
$= 2 - 3 + x^2 = x^2 - 1$
The domain of $g \circ f$ is all real numbers, $(-\infty, \infty)$ , since $f(x)$ is defined for all $x$ .

STEP 5

Evaluate $f \circ h$ .
$f \circ h(x) = f(h(x)) = f\left(\frac{1}{x}\right) = 3 - \left(\frac{1}{x}\right)^2$
The domain of $f \circ h$ is $(0, \infty)$ , since $h(x)$ is defined for $x > 0$ .

STEP 6

Evaluate $g \circ h$ .
$g \circ h(x) = g(h(x)) = g\left(\frac{1}{x}\right) = 2 - \frac{1}{x}$
The domain of $g \circ h$ is $(0, \infty)$ , since $h(x)$ is defined for $x > 0$ .

STEP 7

Evaluate $h \circ f$ .
$h \circ f(x) = h(f(x)) = h(3-x^2) = \frac{1}{3-x^2}$
The domain of $h \circ f$ is determined by $3-x^2 > 0$ , which simplifies to $-\sqrt{3} < x < \sqrt{3}$ .

STEP 8

Evaluate $h \circ g$ .
$h \circ g(x) = h(g(x)) = h(2-x) = \frac{1}{2-x}$
The domain of $h \circ g$ is $x \neq 2$ , or $(-\infty, 2) \cup (2, \infty)$ .

STEP 9

Find the inverse of $f(x) = 4 - 5x$ .
Set $y = 4 - 5x$ and solve for $x$ :
$y = 4 - 5x$ $5x = 4 - y$ $x = \frac{4 - y}{5}$
Thus, the inverse is $f^{-1}(x) = \frac{4 - x}{5}$ .

STEP 10

STEP 11

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Studdy Solution

STEP 1

1. The compositions of functions involve substituting one function into another.2. The domain of a composition is determined by the domain of the inner function and the resulting domain after substitution.3. Inverses of functions are found by solving the equation y=f(x) y = f(x) y=f(x) for x x x.

STEP 2

STEP 3

STEP 4

STEP 5

STEP 6

Evaluate g∘h g \circ h g∘h.g∘h(x)=g(h(x))=g(1x)=2−1x g \circ h(x) = g(h(x)) = g\left(\frac{1}{x}\right) = 2 - \frac{1}{x} g∘h(x)=g(h(x))=g(x1​)=2−x1​The domain of g∘h g \circ h g∘h is (0,∞) (0, \infty) (0,∞), since h(x) h(x) h(x) is defined for x>0 x > 0 x>0.

STEP 7

STEP 8

Evaluate h∘g h \circ g h∘g.h∘g(x)=h(g(x))=h(2−x)=12−x h \circ g(x) = h(g(x)) = h(2-x) = \frac{1}{2-x} h∘g(x)=h(g(x))=h(2−x)=2−x1​The domain of h∘g h \circ g h∘g is x≠2 x \neq 2 x=2, or (−∞,2)∪(2,∞) (-\infty, 2) \cup (2, \infty) (−∞,2)∪(2,∞).

STEP 9

Find the inverse of f(x)=4−5x f(x) = 4 - 5x f(x)=4−5x.Set y=4−5x y = 4 - 5x y=4−5x and solve for x x x:y=4−5x y = 4 - 5x y=4−5x 5x=4−y 5x = 4 - y 5x=4−y x=4−y5 x = \frac{4 - y}{5} x=54−y​Thus, the inverse is f−1(x)=4−x5 f^{-1}(x) = \frac{4 - x}{5} f−1(x)=54−x​.

STEP 10

STEP 11

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1. The compositions of functions involve substituting one function into another.
2. The domain of a composition is determined by the domain of the inner function and the resulting domain after substitution.
3. Inverses of functions are found by solving the equation $y = f(x)$ for $x$ .

Evaluate $g \circ h$ .
$g \circ h(x) = g(h(x)) = g\left(\frac{1}{x}\right) = 2 - \frac{1}{x}$
The domain of $g \circ h$ is $(0, \infty)$ , since $h(x)$ is defined for $x > 0$ .

Evaluate $h \circ g$ .
$h \circ g(x) = h(g(x)) = h(2-x) = \frac{1}{2-x}$
The domain of $h \circ g$ is $x \neq 2$ , or $(-\infty, 2) \cup (2, \infty)$ .

Find the inverse of $f(x) = 4 - 5x$ .
Set $y = 4 - 5x$ and solve for $x$ :
$y = 4 - 5x$ $5x = 4 - y$ $x = \frac{4 - y}{5}$
Thus, the inverse is $f^{-1}(x) = \frac{4 - x}{5}$ .