Question

in exercises 13–26, evaluate the indicated function f(x) = x² − 1 and g(x) = x − 2 algebraically. if pos use a graphing utility to verify your answer. 13. (f + g)(3) 14. (f − g)(−2) 15. (f − g)(0) 16. (f + g)(1)

Explanation:

Let's solve each of these problems one by one. We'll start with problem 13: \((f + g)(3)\)

Problem 13: \((f + g)(3)\)

Step 1: Recall the definition of \((f + g)(x)\)

The sum of two functions \(f(x)\) and \(g(x)\) is defined as \((f + g)(x)=f(x)+g(x)\). So first, we need to find \(f(x)+g(x)\) and then substitute \(x = 3\).

Given \(f(x)=x^{2}-1\) and \(g(x)=x - 2\), then:
\(f(x)+g(x)=(x^{2}-1)+(x - 2)\)
Simplify the right - hand side:
\(f(x)+g(x)=x^{2}+x-1 - 2=x^{2}+x - 3\)

Step 2: Substitute \(x = 3\) into \((f + g)(x)\)

Now we substitute \(x = 3\) into \(x^{2}+x - 3\):
\((f + g)(3)=(3)^{2}+3-3\)
First, calculate \((3)^{2}=9\), then:
\((f + g)(3)=9 + 3-3=9\)

Problem 14: \((f - g)(-2)\)

Step 1: Recall the definition of \((f - g)(x)\)

The difference of two functions \(f(x)\) and \(g(x)\) is defined as \((f - g)(x)=f(x)-g(x)\). So we first find \(f(x)-g(x)\) and then substitute \(x=-2\).

Given \(f(x)=x^{2}-1\) and \(g(x)=x - 2\), then:
\(f(x)-g(x)=(x^{2}-1)-(x - 2)\)
Distribute the negative sign:
\(f(x)-g(x)=x^{2}-1-x + 2\)
Simplify the right - hand side:
\(f(x)-g(x)=x^{2}-x+1\)

Step 2: Substitute \(x=-2\) into \((f - g)(x)\)

Substitute \(x = - 2\) into \(x^{2}-x + 1\):
\((f - g)(-2)=(-2)^{2}-(-2)+1\)
Calculate \((-2)^{2}=4\) and \(-(-2)=2\), then:
\((f - g)(-2)=4 + 2+1=7\)

Problem 15: \((f - g)(0)\)

Step 1: Use the definition of \((f - g)(x)\)

We already know from problem 14 that \((f - g)(x)=x^{2}-x + 1\) (since \((f - g)(x)=f(x)-g(x)=(x^{2}-1)-(x - 2)=x^{2}-x + 1\))

Step 2: Substitute \(x = 0\) into \((f - g)(x)\)

Substitute \(x = 0\) into \(x^{2}-x + 1\):
\((f - g)(0)=(0)^{2}-0 + 1\)
\((f - g)(0)=0-0 + 1=1\)

Problem 16: \((f + g)(1)\)

Answer:

s:

1. \(\boldsymbol{9}\)
2. \(\boldsymbol{7}\)
3. \(\boldsymbol{1}\)
4. \(\boldsymbol{-1}\)