Question

7. Defining $h(x) = x - 2$, $f(x)$ as: $f(x) = \begin{cases} 3 & \text{if } x < 1\\ x - 2 & \text{if } 1 \le x < 3\\ 1 & \text{if } 3 \le x \end{cases}$ and $g(x)$ as: $g(x) = \begin{cases} x^2 - 3 & \text{if } x < 0\\ \lfloor x \rfloor & \text{if } 0 \le x \end{cases}$ (a) evaluate $f(x)$, $g(x)$ and $h(x)$ for $x = -1, 0, 1, 2, 3$ and $4$. (b) evaluate $f(g(1))$, $f(h(1))$, $h(f(1))$, $f(f(2))$, $g(g(3.5))$. (c) graph $f(x)$, $g(x)$ and $h(x)$ for $-5 \le x \le 5$.

          7. Defining $h(x) = x - 2$, $f(x)$ as:

$f(x) = \begin{cases} 3 & \text{if } x < 1\\ x - 2 & \text{if } 1 \le x < 3\\ 1 & \text{if } 3 \le x \end{cases}$
and $g(x)$ as:

$g(x) = \begin{cases} x^2 - 3 & \text{if } x < 0\\ \lfloor x \rfloor & \text{if } 0 \le x \end{cases}$
(a) evaluate $f(x)$, $g(x)$ and $h(x)$ for $x = -1, 0, 1, 2, 3$ and $4$.
(b) evaluate $f(g(1))$, $f(h(1))$, $h(f(1))$, $f(f(2))$, $g(g(3.5))$.
(c) graph $f(x)$, $g(x)$ and $h(x)$ for $-5 \le x \le 5$.

7. Defining h(x) = x - 2, f(x) as:

f(x) = 3 if x < 1
x - 2 if 1 ≤ x < 3
1 if 3 ≤ x
and g(x) as:

g(x) = x^2 - 3 if x < 0
⌊ x ⌋ if 0 ≤ x
(a) evaluate f(x), g(x) and h(x) for x = -1, 0, 1, 2, 3 and 4.
(b) evaluate f(g(1)), f(h(1)), h(f(1)), f(f(2)), g(g(3.5)).
(c) graph f(x), g(x) and h(x) for -5 ≤ x ≤ 5.

Added by Xavier G.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Donna Densmore

12/03/2022

Step 1

To evaluate f(g(x)), we need to substitute the value of g(x) into f(x). For g(x): g(x) = 2 if x ≤ 0 g(x) = 0 if 0 < x For x = -10: g(-10) = 2 (since -10 ≤ 0) f(g(-10)) = f(2) = 2 (since 1 < 2 ≤ 3) For x = -1: g(-1) = 2 (since -1 ≤ 0) f(g(-1)) = f(2) = 2 (since Show more…

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Texts: 7. Defining h(x): 2 if x ≤ 1 f(x): 2 if 1 < x ≤ 3 if 3 < x and g(x): 2 if x ≤ 0 if 0 < x a) Evaluate f(g(x)) and h(g(x)) for x = -10, -1, 2, 3, and 4. b) Evaluate f(f(1)), h(f(f(2))), and g(g(3.5)). c) Graph f(g(x)) and h(g(x)) for -5 ≤ x ≤ 5.