Answer :
To determine where the piecewise function [tex]\( f(x) \)[/tex] is decreasing, we need to analyze the derivative of each segment of the function. A function is decreasing where its derivative is negative.
Given the piecewise function:
[tex]\[ f(x)=\left\{\begin{array}{ll} -(3 x + 7) & \text{for } x < -3 \\ 2 x^2 - 16 & \text{for } -3 \leq x \leq 3 \\ -\left(2^x - 10\right) & \text{for } x > 3 \end{array}\right. \][/tex]
1. For [tex]\( x < -3 \)[/tex]:
[tex]\[ f(x) = -(3x + 7) \][/tex]
Taking the derivative:
[tex]\[ f'(x) = -3 \][/tex]
Since this derivative is a constant and equal to -3 (which is negative), the function is decreasing for [tex]\( x < -3 \)[/tex].
2. For [tex]\( -3 \leq x \leq 3 \)[/tex]:
[tex]\[ f(x) = 2x^2 - 16 \][/tex]
Taking the derivative:
[tex]\[ f'(x) = 4x \][/tex]
The function is decreasing where [tex]\( 4x < 0 \)[/tex]. Solving for [tex]\( x \)[/tex]:
[tex]\[ 4x < 0 \implies x < 0 \][/tex]
Hence, within the range [tex]\( -3 \leq x \leq 3 \)[/tex], the function is decreasing for [tex]\( -3 \leq x < 0 \)[/tex].
3. For [tex]\( x > 3 \)[/tex]:
[tex]\[ f(x) = -\left(2^x - 10\right) \][/tex]
Taking the derivative:
[tex]\[ f'(x) = -2^x \ln(2) \][/tex]
Since [tex]\( 2^x \)[/tex] and [tex]\( \ln(2) \)[/tex] are both positive, their product is positive, making [tex]\( -2^x \ln(2) \)[/tex] negative. Thus, the function is decreasing for [tex]\( x > 3 \)[/tex].
Summary of decreasing sections:
- For [tex]\( x < -3 \)[/tex]
- For [tex]\( -3 \leq x < 0 \)[/tex]
- For [tex]\( x > 3 \)[/tex]
Based on the given piecewise function and its derivative, the sections of the graph where the function is decreasing are: [tex]\( x \in (-\infty, -3) \cup [-3, 0) \cup (3, \infty) \)[/tex].
Combining our identified segments, we have the decreasing sections as:
[tex]\[ x \in \{ -4, -3, -2, -1, 4\} \][/tex]
This matches the result of [tex]\( \{-3, -2, -1, -4, 4\} \)[/tex].
Given the piecewise function:
[tex]\[ f(x)=\left\{\begin{array}{ll} -(3 x + 7) & \text{for } x < -3 \\ 2 x^2 - 16 & \text{for } -3 \leq x \leq 3 \\ -\left(2^x - 10\right) & \text{for } x > 3 \end{array}\right. \][/tex]
1. For [tex]\( x < -3 \)[/tex]:
[tex]\[ f(x) = -(3x + 7) \][/tex]
Taking the derivative:
[tex]\[ f'(x) = -3 \][/tex]
Since this derivative is a constant and equal to -3 (which is negative), the function is decreasing for [tex]\( x < -3 \)[/tex].
2. For [tex]\( -3 \leq x \leq 3 \)[/tex]:
[tex]\[ f(x) = 2x^2 - 16 \][/tex]
Taking the derivative:
[tex]\[ f'(x) = 4x \][/tex]
The function is decreasing where [tex]\( 4x < 0 \)[/tex]. Solving for [tex]\( x \)[/tex]:
[tex]\[ 4x < 0 \implies x < 0 \][/tex]
Hence, within the range [tex]\( -3 \leq x \leq 3 \)[/tex], the function is decreasing for [tex]\( -3 \leq x < 0 \)[/tex].
3. For [tex]\( x > 3 \)[/tex]:
[tex]\[ f(x) = -\left(2^x - 10\right) \][/tex]
Taking the derivative:
[tex]\[ f'(x) = -2^x \ln(2) \][/tex]
Since [tex]\( 2^x \)[/tex] and [tex]\( \ln(2) \)[/tex] are both positive, their product is positive, making [tex]\( -2^x \ln(2) \)[/tex] negative. Thus, the function is decreasing for [tex]\( x > 3 \)[/tex].
Summary of decreasing sections:
- For [tex]\( x < -3 \)[/tex]
- For [tex]\( -3 \leq x < 0 \)[/tex]
- For [tex]\( x > 3 \)[/tex]
Based on the given piecewise function and its derivative, the sections of the graph where the function is decreasing are: [tex]\( x \in (-\infty, -3) \cup [-3, 0) \cup (3, \infty) \)[/tex].
Combining our identified segments, we have the decreasing sections as:
[tex]\[ x \in \{ -4, -3, -2, -1, 4\} \][/tex]
This matches the result of [tex]\( \{-3, -2, -1, -4, 4\} \)[/tex].
