Question

For the following exercises, evaluate the function f at the indicated values x= -3, 0, 2.

Answer 1

10.

• f(-3) = -11
• f(0) = -5
• f(2) = -1

11.

• f(-3) = -1.1
• f(0) = -0.5
• f(2) = -0.1

12.

• f(-3) = 4
• f(0) = -5
• f(2) = -1

Task :

• To evaluate the function f at the indicated values x= -3, 0, 2.

Solution :

In order to evaluate the function f at the indicated values, we simply need to plug in the value of x in the function.

10.

• 
  `f(x) = 2x - 5`

when x = - 3

• 
  `f( - 3) =( 2 * - 3) - 5 \\ f( - 3) = - 6 - 5 \\ f( - 3) = - 11`
• Thus, f(-3) = -11

when x = 0

• 
  `f(0) = 2(0) - 5 \\ f(0) = 0 - 5 \\ f(0) = - 5`
• Thus,f(0) = -5

when x = 2

• 
  `f(2) = (2 * 2)- 5 \\ f(2) = 4 - 5 \\ f(2) = - 1`
• Thus, f(2) = -1

11.

• 
  `f(x) = ( 2x - 5)/(10) \\`

when x = -3

• 
  `f( - 3) = (( 2 * - 3) - 5)/(10) \\f( - 3) = ( - 11)/(10) \\ \\ f( - 3) = - 1.1`
• Thus, f(-3) = -1.1

when x = 0

• 
  `f(0) = ( (2 * 0) - 5)/(10) \\ f(0) = ( 0 - 5)/(10) \\ f( 0) = ( - 5)/(10) \\ f(0) = - 0.5`
• Thus,f(0) = -0.5

when x = 2

• 
  `f(2) = ( (2 * 2) - 5)/(10) \\ f(2) = ( - 1)/(10) \\ f(2) = - 0.1`
• Thus, f(2) = -0.1

12.

• 
  `f(x) = x {}^(2) - 5`

when x = -3

• 
  `f( - 3) = {( - 3)}^(2) - 5 \\ f( - 3) = 9 - 5 \\ f( - 3) = 4`
• Thus, f(-3) = 4

when x = 0

• 
  `f(0) = 0 {}^(2) - 5 \\ f(0) = 0 - 5 \\ f(0) = - 5`
• Thus, f(0) = -5

when x = 2

• 
  `f(2) = (2) {}^(2) - 5 \\ f(2) = 4 - 5 \\ f(2) = - 1`
• Thus, f(2) = -1

Answer 2

`\begin{aligned}\textsf{10)}\quad &f(-3) = -11\\&f(0) = -5\\&f(2) = -1\end{aligned}`

`\begin{aligned}\text{11)}\quad &f(-3) = (11)/(10)\\&f(0) = -(1)/(2)\\&f(2) = -(1)/(10)\end{aligned}`

`\begin{aligned}\text{12)}\quad &f(-3) = 4\\&f(0) = -5\\&f(2) = -1\end{aligned}`

Explanation:

To evaluate a function f(x) at a specific value of x, we need to replace the variable x with the indicated value in the function, then perform any necessary calculations to find the result.

`\hrulefill`

Question 10

Evaluate f(x) = 2x - 5 at x = -3, x = 0 and x = 2.

`\begin{aligned}x=-3 \implies f(-3)&=2(-3)-5\\&=-6-5\\&=-11\end{aligned}`

`\begin{aligned}x=0 \implies f(0)&=2(0)-5\\&=0-5\\&=-5\end{aligned}`

`\begin{aligned}x=2 \implies f(2)&=2(2)-5\\&=4-5\\&=-1\end{aligned}`

Therefore:

• f(-3) = -11
• f(0) = -5
• f(2) = -1

`\hrulefill`

Question 11

Evaluate f(x) = (2x - 5)/10 at x = -3, x = 0 and x = 2.

`\begin{aligned}x=-3 \implies f(-3)&=(2(-3)-5)/(10)\\\\&=(-6-5)/(10)\\\\&=-(11)/(10)\end{aligned}`

`\begin{aligned}x=0 \implies f(0)&=(2(0)-5)/(10)\\\\&=(0-5)/(10)\\\\&=(-5)/(10)\\\\&=-(1)/(2)\end{aligned}`

`\begin{aligned}x=2 \implies f(2)&=(2(2)-5)/(10)\\\\&=(4-5)/(10)\\\\&=-(1)/(10)\end{aligned}`

Therefore:

• f(-3) = 11/10
• f(0) = -1/2
• f(2) = -1/10

`\hrulefill`

Question 12

Evaluate f(x) = x² - 5 at x = -3, x = 0 and x = 2.

`\begin{aligned}x=-3 \implies f(-3)&=(-3)^2-5\\&=9-5\\&=4\end{aligned}`

`\begin{aligned}x=0 \implies f(0)&=(0)^2-5\\&=0-5\\&=-5\end{aligned}`

`\begin{aligned}x=2 \implies f(2)&=(2)^2-5\\&=4-5\\&=-1\end{aligned}`

Therefore:

• f(-3) = 4
• f(0) = -5
• f(2) = -1