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Sagot :
To determine whether the function [tex]\( g(x) = \frac{3 - x}{3^{-x}} \)[/tex] is a growth or decay function on specific intervals, we need to examine the behavior of its first derivative.
### Step-by-Step Solution:
1. Rewrite the function:
We start by expressing the given function in a more convenient form:
[tex]\[ g(x) = \frac{3 - x}{3^{-x}} = (3 - x) \cdot 3^x \][/tex]
This simplifies our analyses going forward.
2. Find the derivative of [tex]\( g(x) \)[/tex]:
To determine the nature (growth or decay) of [tex]\( g(x) \)[/tex], we find the first derivative [tex]\( g'(x) \)[/tex].
Applying the product rule, where [tex]\( u = 3 - x \)[/tex] and [tex]\( v = 3^x \)[/tex], we have:
[tex]\[ g(x) = u(x) \cdot v(x) = (3 - x) \cdot 3^x \][/tex]
The derivative of [tex]\( u(x) \)[/tex] is [tex]\( u'(x) = -1 \)[/tex], and the derivative of [tex]\( v(x) \)[/tex] is [tex]\( v'(x) = 3^x \ln(3) \)[/tex].
Using the product rule:
[tex]\[ g'(x) = u'(x) v(x) + u(x) v'(x) \][/tex]
Substituting in the expressions for [tex]\( u, u', v, \)[/tex] and [tex]\( v' \)[/tex]:
[tex]\[ g'(x) = (-1) \cdot 3^x + (3 - x) \cdot 3^x \ln(3) \][/tex]
Simplifying further:
[tex]\[ g'(x) = -3^x + (3 - x) 3^x \ln(3) \][/tex]
Factoring out [tex]\( 3^x \)[/tex]:
[tex]\[ g'(x) = 3^x [ (3 - x) \ln(3) - 1 ] \][/tex]
3. Evaluate the derivative at specific points:
To understand the behavior of [tex]\( g(x) \)[/tex], we need to test the sign of the derivative in the intervals [tex]\( x < 2 \)[/tex] and [tex]\( x > 2 \)[/tex].
* For [tex]\( x < 2 \)[/tex]:
Let's evaluate [tex]\( g'(x) \)[/tex] at [tex]\( x=1 \)[/tex]:
[tex]\[ g'(1) = 3^1 [ (3 - 1) \ln(3) - 1 ] \][/tex]
Simplify inside the brackets:
[tex]\[ g'(1) = 3 [ 2 \ln(3) - 1 ] \][/tex]
We know from logarithmic properties that [tex]\( \ln(3) \)[/tex] is approximately 1.0986, thus:
[tex]\[ 2 \ln(3) - 1 \approx 2 \cdot 1.0986 - 1 \approx 2.1972 - 1 = 1.1972 \][/tex]
Therefore,
[tex]\[ g'(1) \approx 3 \cdot 1.1972 = 3.5916 \][/tex]
Since [tex]\( g'(1) > 0 \)[/tex], the function is increasing for [tex]\( x < 2 \)[/tex], indicating it is a growth function in this interval.
* For [tex]\( x > 2 \)[/tex]:
Let's evaluate [tex]\( g'(x) \)[/tex] at [tex]\( x=3 \)[/tex]:
[tex]\[ g'(3) = 3^3 [ (3 - 3) \ln(3) - 1 ] = 3^3 [ 0 \cdot \ln(3) - 1 ] = 3^3 (-1) \][/tex]
Simplify:
[tex]\[ g'(3) = -3^3 = -27 \][/tex]
Since [tex]\( g'(3) < 0 \)[/tex], the function is decreasing for [tex]\( x > 2 \)[/tex], indicating it is a decay function in this interval.
### Conclusion:
The function [tex]\( g(x) = \frac{3 - x}{3^{-x}} = (3-x)3^x \)[/tex] is a growth function for [tex]\( x < 2 \)[/tex] and a decay function for [tex]\( x > 2 \)[/tex].
### Step-by-Step Solution:
1. Rewrite the function:
We start by expressing the given function in a more convenient form:
[tex]\[ g(x) = \frac{3 - x}{3^{-x}} = (3 - x) \cdot 3^x \][/tex]
This simplifies our analyses going forward.
2. Find the derivative of [tex]\( g(x) \)[/tex]:
To determine the nature (growth or decay) of [tex]\( g(x) \)[/tex], we find the first derivative [tex]\( g'(x) \)[/tex].
Applying the product rule, where [tex]\( u = 3 - x \)[/tex] and [tex]\( v = 3^x \)[/tex], we have:
[tex]\[ g(x) = u(x) \cdot v(x) = (3 - x) \cdot 3^x \][/tex]
The derivative of [tex]\( u(x) \)[/tex] is [tex]\( u'(x) = -1 \)[/tex], and the derivative of [tex]\( v(x) \)[/tex] is [tex]\( v'(x) = 3^x \ln(3) \)[/tex].
Using the product rule:
[tex]\[ g'(x) = u'(x) v(x) + u(x) v'(x) \][/tex]
Substituting in the expressions for [tex]\( u, u', v, \)[/tex] and [tex]\( v' \)[/tex]:
[tex]\[ g'(x) = (-1) \cdot 3^x + (3 - x) \cdot 3^x \ln(3) \][/tex]
Simplifying further:
[tex]\[ g'(x) = -3^x + (3 - x) 3^x \ln(3) \][/tex]
Factoring out [tex]\( 3^x \)[/tex]:
[tex]\[ g'(x) = 3^x [ (3 - x) \ln(3) - 1 ] \][/tex]
3. Evaluate the derivative at specific points:
To understand the behavior of [tex]\( g(x) \)[/tex], we need to test the sign of the derivative in the intervals [tex]\( x < 2 \)[/tex] and [tex]\( x > 2 \)[/tex].
* For [tex]\( x < 2 \)[/tex]:
Let's evaluate [tex]\( g'(x) \)[/tex] at [tex]\( x=1 \)[/tex]:
[tex]\[ g'(1) = 3^1 [ (3 - 1) \ln(3) - 1 ] \][/tex]
Simplify inside the brackets:
[tex]\[ g'(1) = 3 [ 2 \ln(3) - 1 ] \][/tex]
We know from logarithmic properties that [tex]\( \ln(3) \)[/tex] is approximately 1.0986, thus:
[tex]\[ 2 \ln(3) - 1 \approx 2 \cdot 1.0986 - 1 \approx 2.1972 - 1 = 1.1972 \][/tex]
Therefore,
[tex]\[ g'(1) \approx 3 \cdot 1.1972 = 3.5916 \][/tex]
Since [tex]\( g'(1) > 0 \)[/tex], the function is increasing for [tex]\( x < 2 \)[/tex], indicating it is a growth function in this interval.
* For [tex]\( x > 2 \)[/tex]:
Let's evaluate [tex]\( g'(x) \)[/tex] at [tex]\( x=3 \)[/tex]:
[tex]\[ g'(3) = 3^3 [ (3 - 3) \ln(3) - 1 ] = 3^3 [ 0 \cdot \ln(3) - 1 ] = 3^3 (-1) \][/tex]
Simplify:
[tex]\[ g'(3) = -3^3 = -27 \][/tex]
Since [tex]\( g'(3) < 0 \)[/tex], the function is decreasing for [tex]\( x > 2 \)[/tex], indicating it is a decay function in this interval.
### Conclusion:
The function [tex]\( g(x) = \frac{3 - x}{3^{-x}} = (3-x)3^x \)[/tex] is a growth function for [tex]\( x < 2 \)[/tex] and a decay function for [tex]\( x > 2 \)[/tex].
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