**Mathematics Practice Problems****a) Problem:**Calculate the value of $ f(g(h(3))) $.Given functions:- $ f(x) = 2^x $- $ g(x) = x^2 $- $ h(x) = \frac{x}{4} $**b) Problem:**Prove that the function $ f: \mathbb{R} \rightarrow \mathbb{R} $ defined by $ f(x) = |x| $ is not surjective and is not injective.---**Solutions:****a) Solution:**1. **Find $ h(3) $:** \[ h(3) = \frac{3}{4} \]2. **Find $ g(h(3)) $:** \[ g\left(\frac{3}{4}\right) = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \]3. **Find $ f(g(h(3))) $:** \[ f\left(\frac{9}{16}\right) = 2^{\frac{9}{16}} \]**b) Solution:***To prove $ f: \mathbb{R} \to \mathbb{R} $, $ f(x) = |x| $ is not surjective:*- A function is surjective if for every $ y $ in the codomain, there is an $ x $ in the domain such that $ f(x) = y $.- The codomain is $\mathbb{R}$, but $ f(x) = |x| $ only outputs non-negative values. Therefore, negative numbers in $\mathbb{R}$ have no pre-image under $ f $. - Thus, $ f $ is not surjective.*To prove $ f: \mathbb{R} \to \mathbb{R} $, $ f(x) = |x| $ is not injective:*- A function is injective if $ f(x_1) = f(x_2) $ implies $ x_1 = x_2 $.- For instance, $ f(1) = 1 $ and $ f(-1) = 1 $, so

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Answered: a)-What is fof g ofh 3)? fari= gal=…

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a)-What is fof g ofh 3)? fari= gal= bprove f:R¬R fer )=4lis not Surgective and is not injective.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

**Mathematics Practice Problems**

**a) Problem:**
Calculate the value of $ f(g(h(3))) $.

Given functions:
- $ f(x) = 2^x $
- $ g(x) = x^2 $
- $ h(x) = \frac{x}{4} $

**b) Problem:**
Prove that the function $ f: \mathbb{R} \rightarrow \mathbb{R} $ defined by $ f(x) = |x| $ is not surjective and is not injective.

---

**Solutions:**

**a) Solution:**

1. **Find $ h(3) $:**
\[
h(3) = \frac{3}{4}
\]

2. **Find $ g(h(3)) $:**
\[
g\left(\frac{3}{4}\right) = \left(\frac{3}{4}\right)^2 = \frac{9}{16}
\]

3. **Find $ f(g(h(3))) $:**
\[
f\left(\frac{9}{16}\right) = 2^{\frac{9}{16}}
\]

**b) Solution:**

*To prove $ f: \mathbb{R} \to \mathbb{R} $, $ f(x) = |x| $ is not surjective:*

- A function is surjective if for every $ y $ in the codomain, there is an $ x $ in the domain such that $ f(x) = y $.
- The codomain is $\mathbb{R}$, but $ f(x) = |x| $ only outputs non-negative values. Therefore, negative numbers in $\mathbb{R}$ have no pre-image under $ f $.
- Thus, $ f $ is not surjective.

*To prove $ f: \mathbb{R} \to \mathbb{R} $, $ f(x) = |x| $ is not injective:*

- A function is injective if $ f(x_1) = f(x_2) $ implies $ x_1 = x_2 $.
- For instance, $ f(1) = 1 $ and $ f(-1) = 1 $, so

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