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Sagot :
To evaluate the piece-wise function for the given values of [tex]\( x \)[/tex], we will apply the appropriate piece of the function based on the value of [tex]\( x \)[/tex]. The function [tex]\( f(x) \)[/tex] is defined as follows:
[tex]\[ f(x)=\left\{\begin{array}{cc} -x^2+2 & \text{if } x \leq -3 \\ \frac{4-|x|}{\sqrt[3]{x+1}} & \text{if } -1 \leq x \leq 1 \\ x & \text{if } x>1 \end{array}\right. \][/tex]
We need to evaluate this function for the values [tex]\( x = -5, -3, -1, 0, 1, 2 \)[/tex]:
1. For [tex]\( x = -5 \)[/tex]:
- Since [tex]\( x \leq -3 \)[/tex], we use the piece [tex]\( -x^2 + 2 \)[/tex].
- [tex]\( f(-5) = -(-5)^2 + 2 = -25 + 2 = -23 \)[/tex].
2. For [tex]\( x = -3 \)[/tex]:
- Since [tex]\( x \leq -3 \)[/tex], we use the piece [tex]\( -x^2 + 2 \)[/tex].
- [tex]\( f(-3) = -(-3)^2 + 2 = -9 + 2 = -7 \)[/tex].
3. For [tex]\( x = -1 \)[/tex]:
- Since [tex]\( -1 \leq x \leq 1 \)[/tex], we use the piece [tex]\( \frac{4 - |x|}{\sqrt[3]{x + 1}} \)[/tex].
- First, evaluate if the denominator becomes zero: [tex]\(\sqrt[3]{x + 1} = \sqrt[3]{-1 + 1} = \sqrt[3]{0}\)[/tex], which is undefined.
- Therefore, [tex]\( f(-1) \)[/tex] is undefined.
4. For [tex]\( x = 0 \)[/tex]:
- Since [tex]\( -1 \leq x \leq 1 \)[/tex], we use the piece [tex]\( \frac{4 - |x|}{\sqrt[3]{x + 1}} \)[/tex].
- Evaluate the expression:
[tex]\[ f(0) = \frac{4 - |0|}{\sqrt[3]{0 + 1}} = \frac{4 - 0}{\sqrt[3]{1}} = \frac{4}{1} = 4 \][/tex]
5. For [tex]\( x = 1 \)[/tex]:
- Since [tex]\( -1 \leq x \leq 1 \)[/tex], we use the piece [tex]\( \frac{4 - |x|}{\sqrt[3]{x + 1}} \)[/tex].
- Evaluate the expression:
[tex]\[ f(1) = \frac{4 - |1|}{\sqrt[3]{1 + 1}} = \frac{4 - 1}{\sqrt[3]{2}} = \frac{3}{\sqrt[3]{2}} \][/tex]
Numerically, [tex]\( \sqrt[3]{2} \approx 1.2599 \)[/tex], so:
[tex]\[ f(1) \approx \frac{3}{1.2599} \approx 2.381 \][/tex]
6. For [tex]\( x = 2 \)[/tex]:
- Since [tex]\( x > 1 \)[/tex], we use the piece [tex]\( x \)[/tex].
- [tex]\( f(2) = 2 \)[/tex].
So, summarizing the results:
- [tex]\( f(-5) = -23 \)[/tex]
- [tex]\( f(-3) = -7 \)[/tex]
- [tex]\( f(-1) = \text{undefined} \)[/tex]
- [tex]\( f(0) = 4 \)[/tex]
- [tex]\( f(1) \approx 2.381 \)[/tex]
- [tex]\( f(2) = 2 \)[/tex]
Thus, the evaluated values of [tex]\( f(x) \)[/tex] for the given [tex]\( x \)[/tex] values are:
[tex]\[ [-23, -7, \text{undefined}, 4, 2.381, 2] \][/tex]
[tex]\[ f(x)=\left\{\begin{array}{cc} -x^2+2 & \text{if } x \leq -3 \\ \frac{4-|x|}{\sqrt[3]{x+1}} & \text{if } -1 \leq x \leq 1 \\ x & \text{if } x>1 \end{array}\right. \][/tex]
We need to evaluate this function for the values [tex]\( x = -5, -3, -1, 0, 1, 2 \)[/tex]:
1. For [tex]\( x = -5 \)[/tex]:
- Since [tex]\( x \leq -3 \)[/tex], we use the piece [tex]\( -x^2 + 2 \)[/tex].
- [tex]\( f(-5) = -(-5)^2 + 2 = -25 + 2 = -23 \)[/tex].
2. For [tex]\( x = -3 \)[/tex]:
- Since [tex]\( x \leq -3 \)[/tex], we use the piece [tex]\( -x^2 + 2 \)[/tex].
- [tex]\( f(-3) = -(-3)^2 + 2 = -9 + 2 = -7 \)[/tex].
3. For [tex]\( x = -1 \)[/tex]:
- Since [tex]\( -1 \leq x \leq 1 \)[/tex], we use the piece [tex]\( \frac{4 - |x|}{\sqrt[3]{x + 1}} \)[/tex].
- First, evaluate if the denominator becomes zero: [tex]\(\sqrt[3]{x + 1} = \sqrt[3]{-1 + 1} = \sqrt[3]{0}\)[/tex], which is undefined.
- Therefore, [tex]\( f(-1) \)[/tex] is undefined.
4. For [tex]\( x = 0 \)[/tex]:
- Since [tex]\( -1 \leq x \leq 1 \)[/tex], we use the piece [tex]\( \frac{4 - |x|}{\sqrt[3]{x + 1}} \)[/tex].
- Evaluate the expression:
[tex]\[ f(0) = \frac{4 - |0|}{\sqrt[3]{0 + 1}} = \frac{4 - 0}{\sqrt[3]{1}} = \frac{4}{1} = 4 \][/tex]
5. For [tex]\( x = 1 \)[/tex]:
- Since [tex]\( -1 \leq x \leq 1 \)[/tex], we use the piece [tex]\( \frac{4 - |x|}{\sqrt[3]{x + 1}} \)[/tex].
- Evaluate the expression:
[tex]\[ f(1) = \frac{4 - |1|}{\sqrt[3]{1 + 1}} = \frac{4 - 1}{\sqrt[3]{2}} = \frac{3}{\sqrt[3]{2}} \][/tex]
Numerically, [tex]\( \sqrt[3]{2} \approx 1.2599 \)[/tex], so:
[tex]\[ f(1) \approx \frac{3}{1.2599} \approx 2.381 \][/tex]
6. For [tex]\( x = 2 \)[/tex]:
- Since [tex]\( x > 1 \)[/tex], we use the piece [tex]\( x \)[/tex].
- [tex]\( f(2) = 2 \)[/tex].
So, summarizing the results:
- [tex]\( f(-5) = -23 \)[/tex]
- [tex]\( f(-3) = -7 \)[/tex]
- [tex]\( f(-1) = \text{undefined} \)[/tex]
- [tex]\( f(0) = 4 \)[/tex]
- [tex]\( f(1) \approx 2.381 \)[/tex]
- [tex]\( f(2) = 2 \)[/tex]
Thus, the evaluated values of [tex]\( f(x) \)[/tex] for the given [tex]\( x \)[/tex] values are:
[tex]\[ [-23, -7, \text{undefined}, 4, 2.381, 2] \][/tex]
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