For the piecewise function, find the specified function value.

[tex]\[
f(x) =
\begin{cases}
9x + 1, & \text{for } x \ \textless \ 8 \\
8x, & \text{for } 8 \leq x \leq 10 \\
8 - 9x, & \text{for } x \ \textgreater \ 10
\end{cases}
\][/tex]

A. -71
B. 80
C. 73
D. -64



Answer :

Certainly! Let's solve this problem step-by-step for each given value of [tex]\(x\)[/tex].

### For [tex]\(x = -71\)[/tex]:

Since [tex]\( -71 < 8 \)[/tex], we will use the piece of the function defined for [tex]\( x < 8 \)[/tex]:

[tex]\[ f(x) = 9x + 1 \][/tex]

Substituting [tex]\( x = -71 \)[/tex], we get:

[tex]\[ f(-71) = 9(-71) + 1 \][/tex]

[tex]\[ = -639 + 1 \][/tex]

[tex]\[ = -638 \][/tex]

So, [tex]\( f(-71) = -638 \)[/tex].

### For [tex]\(x = 80\)[/tex]:

Since [tex]\( 80 > 10 \)[/tex], we will use the piece of the function defined for [tex]\( x > 10 \)[/tex]:

[tex]\[ f(x) = 8 - 9x \][/tex]

Substituting [tex]\( x = 80 \)[/tex], we get:

[tex]\[ f(80) = 8 - 9(80) \][/tex]

[tex]\[ = 8 - 720 \][/tex]

[tex]\[ = -712 \][/tex]

So, [tex]\( f(80) = -712 \)[/tex].

### For [tex]\(x = 73\)[/tex]:

Since [tex]\( 73 > 10 \)[/tex], we will use the piece of the function defined for [tex]\( x > 10 \)[/tex]:

[tex]\[ f(x) = 8 - 9x \][/tex]

Substituting [tex]\( x = 73 \)[/tex], we get:

[tex]\[ f(73) = 8 - 9(73) \][/tex]

[tex]\[ = 8 - 657 \][/tex]

[tex]\[ = -649 \][/tex]

So, [tex]\( f(73) = -649 \)[/tex].

### For [tex]\(x = -64\)[/tex]:

Since [tex]\( -64 < 8 \)[/tex], we will use the piece of the function defined for [tex]\( x < 8 \)[/tex]:

[tex]\[ f(x) = 9x + 1 \][/tex]

Substituting [tex]\( x = -64 \)[/tex], we get:

[tex]\[ f(-64) = 9(-64) + 1 \][/tex]

[tex]\[ = -576 + 1 \][/tex]

[tex]\[ = -575 \][/tex]

So, [tex]\( f(-64) = -575 \)[/tex].

### Summary:

- [tex]\( f(-71) = -638 \)[/tex]
- [tex]\( f(80) = -712 \)[/tex]
- [tex]\( f(73) = -649 \)[/tex]
- [tex]\( f(-64) = -575 \)[/tex]

Therefore, the specified function values are:
- A. -638
- B. -712
- C. -649
- D. -575