Question

1. if \\(f(x) = 3x - 2\\) and \\(g(x) = x^2 + 1\\), find

a.) \\(f(4)\\)
b.) \\(g(-3)\\)
c.) \\(f(2) + g(2)\\)

1. given \\(f(x) = 2x + 5\\) and \\(g(x) = x - 1\\), find

a.) \\((f + g)(x)\\)
b.) \\((f - g)(x)\\)
c.) \\((f \cdot g)(x)\\)

1. if \\(f(x) = x^2 - 4\\) and \\(g(x) = x + 3\\), find \\(\frac{f(x)}{g(x)}\\)

Explanation:

The image is oriented upside down. Let's first transcribe the three questions accurately by rotating the text mentally:

Question 1:
If \(f(x) = 3x - 2\) and \(g(x) = x^2 + 1\), find:
a.) \(f(4)\)
b.) \(g(-3)\)
c.) \(f(2) + g(2)\)

Question 2:
Given \(f(x) = 2x + 5\) and \(g(x) = x - 1\), find:
a.) \((f + g)(x)\)
b.) \((f - g)(x)\)
c.) \((f \cdot g)(x)\)

Question 3:
If \(f(x) = x^2 - 4\) and \(g(x) = x + 3\), find \(\frac{f(x)}{g(x)}\).

Let's solve each question step-by-step.

Evaluate functions for Question 1

We are given \(f(x) = 3x - 2\) and \(g(x) = x^2 + 1\).

For part a.), substitute \(x = 4\) into \(f(x)\):

$$ f(4) = 3(4) - 2 = 12 - 2 = 10 $$

For part b.), substitute \(x = -3\) into \(g(x)\):

$$ g(-3) = (-3)^2 + 1 = 9 + 1 = 10 $$

For part c.), first find \(f(2)\) and \(g(2)\):

$$ f(2) = 3(2) - 2 = 6 - 2 = 4 $$

$$ g(2) = (2)^2 + 1 = 4 + 1 = 5 $$

Then add them together:

$$ f(2) + g(2) = 4 + 5 = 9 $$

Perform operations for Question 2

We are given \(f(x) = 2x + 5\) and \(g(x) = x - 1\).

For part a.), find the sum function:

$$ (f + g)(x) = f(x) + g(x) = (2x + 5) + (x - 1) = 3x + 4 $$

For part b.), find the difference function:

$$ (f - g)(x) = f(x) - g(x) = (2x + 5) - (x - 1) = 2x + 5 - x + 1 = x + 6 $$

For part c.), find the product function:

$$ (f \cdot g)(x) = f(x) \cdot g(x) = (2x + 5)(x - 1) = 2x^2 - 2x + 5x - 5 = 2x^2 + 3x - 5 $$

Find the quotient for Question 3

We are given \(f(x) = x^2 - 4\) and \(g(x) = x + 3\).

Find the quotient function \(\frac{f(x)}{g(x)}\):

$$ \frac{f(x)}{g(x)} = \frac{x^2 - 4}{x + 3} $$

Since \(x^2 - 4 = (x - 2)(x + 2)\), there are no common factors with the denominator \(x + 3\) to simplify further. Thus, the expression is:

$$ \frac{x^2 - 4}{x + 3} \quad (\text{where } x eq -3) $$

Answer:

Question 1

If \(f(x) = 3x - 2\) and \(g(x) = x^2 + 1\):

• a.) \(f(4) = 10\)
• b.) \(g(-3) = 10\)
• c.) \(f(2) + g(2) = 9\)

Question 2

Given \(f(x) = 2x + 5\) and \(g(x) = x - 1\):

• a.) \((f + g)(x) = 3x + 4\)
• b.) \((f - g)(x) = x + 6\)
• c.) \((f \cdot g)(x) = 2x^2 + 3x - 5\)

Question 3

If \(f(x) = x^2 - 4\) and \(g(x) = x + 3\):

$$ \frac{f(x)}{g(x)} = \frac{x^2 - 4}{x + 3} \quad (x eq -3) $$