Question86 FUNCTIONS (Chapter 3) 11 Suppose $f(x)=\sqrt{1-x}$ and $g(x)=x^{2}$ . Find: a $(f \circ g)(x)$ b the domain and range of $(f \circ g)(x)$ .

Studdy Solution

STEP 1

1. We are given two functions: $f(x) = \sqrt{1-x}$ and $g(x) = x^2$ .
2. We need to find the composition $(f \circ g)(x)$ .
3. We need to determine the domain and range of $(f \circ g)(x)$ .

STEP 2

1. Find the expression for $(f \circ g)(x)$ .
2. Determine the domain of $(f \circ g)(x)$ .
3. Determine the range of $(f \circ g)(x)$ .

STEP 3

To find $(f \circ g)(x)$ , substitute $g(x) = x^2$ into $f(x)$ :
$(f \circ g)(x) = f(g(x)) = f(x^2) = \sqrt{1 - x^2}$

STEP 4

Determine the domain of $(f \circ g)(x) = \sqrt{1 - x^2}$ .
The expression under the square root, $1 - x^2$ , must be non-negative:
$1 - x^2 \geq 0$
This implies:
$1 \geq x^2$
Taking square roots:
$-1 \leq x \leq 1$
Thus, the domain of $(f \circ g)(x)$ is:
$[-1, 1]$

STEP 5

Determine the range of $(f \circ g)(x) = \sqrt{1 - x^2}$ .
The expression $\sqrt{1 - x^2}$ represents the upper half of a circle with radius 1 centered at the origin. The values of $\sqrt{1 - x^2}$ range from 0 to 1 as $x$ varies from $-1$ to $1$ .
Thus, the range of $(f \circ g)(x)$ is:
$[0, 1]$
The expression for $(f \circ g)(x)$ is $\sqrt{1 - x^2}$ , with domain $[-1, 1]$ and range $[0, 1]$ .

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