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Chapter 0: Problem 7

If $$f(x)=\left\\{\begin{array}{ll} x^{2}+1 & \text { if } x \leq 0 \\ \sqrt{x} & \text { if } x>0 \end{array}\right.$$ find \(f(-2), f(0)\), and \(f(1)\).

Short Answer

Expert verified
\(f(-2) = 5, f(0) = 1, \) and \(f(1) = 1\)

Step by step solution

01

Determine which part of the function to use

Since -2 is less than 0, we will use the part of the function for x 鈮� 0: \(f(x) = x^2 + 1\)
02

Calculate for x = -2

We will plug in x = -2 into the function to find the output: \(f(-2) = (-2)^2 + 1 = 4 + 1 = 5\) #Case 2#: x = 0
03

Determine which part of the function to use

Since 0 is equal to 0, we will again use the part of the function for x 鈮� 0: \(f(x) = x^2+1\)
04

Calculate for x = 0

We will plug in x = 0 into the function to find the output: \(f(0) = (0)^2+1 = 0+1 = 1\) #Case 3#: x = 1
05

Determine which part of the function to use

Since 1 is greater than 0, we will use the part of the function for x > 0: \(f(x) = \sqrt{x}\)
06

Calculate for x = 1

We will plug in x = 1 into the function to find the output: \(f(1) = \sqrt{1} = 1\) So, we have \(f(-2) = 5, f(0) = 1, \) and \(f(1) = 1\).

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Prove that a function has an inverse if and only if it is oneto-one.

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