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Sagot :
Let's analyze the function \( y = x \cdot 5^x \) and address each part of the question step by step.
### Part (a) - Intervals of Increasing and Decreasing
To determine the intervals on which the function is increasing or decreasing, we need to find the first derivative of the function \( y \) and identify the critical points.
1. First Derivative:
The first derivative of the function \( y = x \cdot 5^x \) is found by using the product rule:
[tex]\[ y' = \frac{d}{dx} \left( x \cdot 5^x \right) = 5^x + x \cdot 5^x \ln(5) \][/tex]
Simplifying further:
[tex]\[ y' = 5^x (1 + x \ln(5)) \][/tex]
2. Critical Points:
To find the critical points, we set the first derivative equal to zero:
[tex]\[ 5^x (1 + x \ln(5)) = 0 \][/tex]
Since \( 5^x \) is never zero, we focus on:
[tex]\[ 1 + x \ln(5) = 0 \implies x = -\frac{1}{\ln(5)} \][/tex]
Thus, the critical point is:
[tex]\[ x = -\frac{1}{\ln(5)} \][/tex]
3. Intervals of Increase and Decreasing:
We need to test intervals around the critical point to determine whether the function is increasing or decreasing.
- For \( x < -\frac{1}{\ln(5)} \):
Choose a point slightly less than \( -\frac{1}{\ln(5)} \).
- If substituted into the derivative \( y' \), the resulting value is negative, showing that the function is decreasing in this interval.
- For \( x > -\frac{1}{\ln(5)} \):
Choose a point slightly greater than \( -\frac{1}{\ln(5)} \).
- If substituted into the derivative \( y' \), the resulting value is positive, showing that the function is increasing in this interval.
Therefore, the intervals are:
- Increasing: \( \left( -\frac{1}{\ln(5)}, \infty \right) \)
- Decreasing: \( \left( -\infty, -\frac{1}{\ln(5)} \right) \)
### Part (b) - Range of the Function
To determine the range of the function, we look at the behavior of \( y = x \cdot 5^x \) as \( x \) approaches positive and negative infinity.
1. As \( x \to +\infty \):
[tex]\[ y = x \cdot 5^x \to \infty \][/tex]
2. As \( x \to -\infty \):
[tex]\[ y = x \cdot 5^x \to 0 \][/tex]
This is because \( 5^x \) approaches \( 0 \) faster than \( x \) approaches \(-\infty\).
Combining these details, the range of the function is:
[tex]\[ \left( 0, \infty \right) \][/tex]
### Final Answers:
(a) Intervals on which the function is:
- Increasing: \( \left( -\frac{1}{\ln(5)}, \infty \right) \)
- Decreasing: \( \left( -\infty, -\frac{1}{\ln(5)} \right) \)
(b) The range of the function:
[tex]\[ (0, \infty) \][/tex]
### Part (a) - Intervals of Increasing and Decreasing
To determine the intervals on which the function is increasing or decreasing, we need to find the first derivative of the function \( y \) and identify the critical points.
1. First Derivative:
The first derivative of the function \( y = x \cdot 5^x \) is found by using the product rule:
[tex]\[ y' = \frac{d}{dx} \left( x \cdot 5^x \right) = 5^x + x \cdot 5^x \ln(5) \][/tex]
Simplifying further:
[tex]\[ y' = 5^x (1 + x \ln(5)) \][/tex]
2. Critical Points:
To find the critical points, we set the first derivative equal to zero:
[tex]\[ 5^x (1 + x \ln(5)) = 0 \][/tex]
Since \( 5^x \) is never zero, we focus on:
[tex]\[ 1 + x \ln(5) = 0 \implies x = -\frac{1}{\ln(5)} \][/tex]
Thus, the critical point is:
[tex]\[ x = -\frac{1}{\ln(5)} \][/tex]
3. Intervals of Increase and Decreasing:
We need to test intervals around the critical point to determine whether the function is increasing or decreasing.
- For \( x < -\frac{1}{\ln(5)} \):
Choose a point slightly less than \( -\frac{1}{\ln(5)} \).
- If substituted into the derivative \( y' \), the resulting value is negative, showing that the function is decreasing in this interval.
- For \( x > -\frac{1}{\ln(5)} \):
Choose a point slightly greater than \( -\frac{1}{\ln(5)} \).
- If substituted into the derivative \( y' \), the resulting value is positive, showing that the function is increasing in this interval.
Therefore, the intervals are:
- Increasing: \( \left( -\frac{1}{\ln(5)}, \infty \right) \)
- Decreasing: \( \left( -\infty, -\frac{1}{\ln(5)} \right) \)
### Part (b) - Range of the Function
To determine the range of the function, we look at the behavior of \( y = x \cdot 5^x \) as \( x \) approaches positive and negative infinity.
1. As \( x \to +\infty \):
[tex]\[ y = x \cdot 5^x \to \infty \][/tex]
2. As \( x \to -\infty \):
[tex]\[ y = x \cdot 5^x \to 0 \][/tex]
This is because \( 5^x \) approaches \( 0 \) faster than \( x \) approaches \(-\infty\).
Combining these details, the range of the function is:
[tex]\[ \left( 0, \infty \right) \][/tex]
### Final Answers:
(a) Intervals on which the function is:
- Increasing: \( \left( -\frac{1}{\ln(5)}, \infty \right) \)
- Decreasing: \( \left( -\infty, -\frac{1}{\ln(5)} \right) \)
(b) The range of the function:
[tex]\[ (0, \infty) \][/tex]
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