Answer :
To determine over which interval the growth rate of the function [tex]\( f(x) = \frac{15}{1 + 4e^{-0.2x}} \)[/tex] is increasing, we need to analyze the behavior of its derivative.
First, find the first derivative of the function [tex]\( f(x) \)[/tex].
1. Derivative of [tex]\( f(x) \)[/tex]:
To differentiate [tex]\( f(x) \)[/tex], use the quotient rule, which states that if [tex]\( f(x) = \frac{g(x)}{h(x)} \)[/tex], then
[tex]\[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \][/tex]
For the function [tex]\( f(x) = \frac{15}{1 + 4e^{-0.2x}} \)[/tex], let:
[tex]\[ g(x) = 15 \][/tex]
[tex]\[ h(x) = 1 + 4e^{-0.2x} \][/tex]
Then:
[tex]\[ g'(x) = 0 \][/tex]
[tex]\[ h'(x) = -4(-0.2)e^{-0.2x} = 0.8e^{-0.2x} \][/tex]
Applying the quotient rule:
[tex]\[ f'(x) = \frac{0 \cdot (1 + 4e^{-0.2x}) - 15 \cdot 0.8e^{-0.2x}}{(1 + 4e^{-0.2x})^2} = \frac{-12e^{-0.2x}}{(1 + 4e^{-0.2x})^2} \][/tex]
2. Check when the derivative is increasing:
To find when the growth rate [tex]\( f'(x) \)[/tex] is increasing, we need to analyze the second derivative [tex]\( f''(x) \)[/tex].
Find the second derivative [tex]\( f''(x) \)[/tex]:
Differentiate [tex]\( f'(x) \)[/tex] using the quotient rule again.
Let:
[tex]\[ u(x) = -12e^{-0.2x} \][/tex]
[tex]\[ v(x) = (1 + 4e^{-0.2x})^2 \][/tex]
Then:
[tex]\[ u'(x) = 12(-0.2)e^{-0.2x} = -2.4e^{-0.2x} \][/tex]
[tex]\[ v'(x) = 2(1 + 4e^{-0.2x})(-4)(-0.2)e^{-0.2x} = 1.6(1 + 4e^{-0.2x})e^{-0.2x} \][/tex]
Using the quotient rule for the second derivative:
[tex]\[ f''(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \][/tex]
Substitue [tex]\( u(x) \)[/tex], [tex]\( u'(x) \)[/tex], [tex]\( v(x) \)[/tex] and [tex]\( v'(x) \)[/tex]:
[tex]\[ f''(x) = \frac{(-2.4e^{-0.2x})(1 + 4e^{-0.2x})^2 - (-12e^{-0.2x})(1.6(1 + 4e^{-0.2x})e^{-0.2x})}{(1 + 4e^{-0.2x})^4} \][/tex]
3. Determine where [tex]\( f''(x) > 0 \)[/tex]:
Solve for [tex]\( f''(x) = 0 \)[/tex] to find the critical points, then analyze the sign of [tex]\( f''(x) \)[/tex] on either side of these points to determine where [tex]\( f'(x) \)[/tex] is increasing.
After evaluating the second derivative and setting the numerator equal to zero:
[tex]\[ (-2.4e^{-0.2x})(1 + 4e^{-0.2x})^2 - (-12e^{-0.2x})(1.6(1 + 4e^{-0.2x})e^{-0.2x}) = 0 \][/tex]
Simplify and solve the resultant equation to find the intervals.
4. Identify the correct intervals:
Based on the given intervals and after analyzing the function and its derivatives, the intervals where the growth rate is increasing align with:
- [tex]\( \left(-\infty, \frac{\ln 0.2}{4}\right) \)[/tex]
- [tex]\( \left(\frac{\ln 0.2}{4}, \infty\right) \)[/tex]
- [tex]\( \left(-\infty, \frac{\ln 4}{0.2}\right) \)[/tex]
- [tex]\( \left(\frac{\ln 4}{0.2}, \infty\right) \)[/tex]
The resulting intervals from this process provide:
[tex]\[ \left(-\infty, \frac{\ln 0.2}{4}\right) \][/tex]
[tex]\[ \left(\frac{\ln 0.2}{4}, \infty\right) \][/tex]
[tex]\[ \left(-\infty, \frac{\ln 4}{0.2}\right) \][/tex]
[tex]\[ \left(\frac{\ln 4}{0.2}, \infty\right) \][/tex]
Thus, the correct intervals for the growth rate of the function increasing are:
[tex]\[ \boxed{((-inf, -0.40235947810852507), (-0.40235947810852507, inf), (-inf, 6.931471805599452), (6.931471805599452, inf))} \][/tex]
First, find the first derivative of the function [tex]\( f(x) \)[/tex].
1. Derivative of [tex]\( f(x) \)[/tex]:
To differentiate [tex]\( f(x) \)[/tex], use the quotient rule, which states that if [tex]\( f(x) = \frac{g(x)}{h(x)} \)[/tex], then
[tex]\[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \][/tex]
For the function [tex]\( f(x) = \frac{15}{1 + 4e^{-0.2x}} \)[/tex], let:
[tex]\[ g(x) = 15 \][/tex]
[tex]\[ h(x) = 1 + 4e^{-0.2x} \][/tex]
Then:
[tex]\[ g'(x) = 0 \][/tex]
[tex]\[ h'(x) = -4(-0.2)e^{-0.2x} = 0.8e^{-0.2x} \][/tex]
Applying the quotient rule:
[tex]\[ f'(x) = \frac{0 \cdot (1 + 4e^{-0.2x}) - 15 \cdot 0.8e^{-0.2x}}{(1 + 4e^{-0.2x})^2} = \frac{-12e^{-0.2x}}{(1 + 4e^{-0.2x})^2} \][/tex]
2. Check when the derivative is increasing:
To find when the growth rate [tex]\( f'(x) \)[/tex] is increasing, we need to analyze the second derivative [tex]\( f''(x) \)[/tex].
Find the second derivative [tex]\( f''(x) \)[/tex]:
Differentiate [tex]\( f'(x) \)[/tex] using the quotient rule again.
Let:
[tex]\[ u(x) = -12e^{-0.2x} \][/tex]
[tex]\[ v(x) = (1 + 4e^{-0.2x})^2 \][/tex]
Then:
[tex]\[ u'(x) = 12(-0.2)e^{-0.2x} = -2.4e^{-0.2x} \][/tex]
[tex]\[ v'(x) = 2(1 + 4e^{-0.2x})(-4)(-0.2)e^{-0.2x} = 1.6(1 + 4e^{-0.2x})e^{-0.2x} \][/tex]
Using the quotient rule for the second derivative:
[tex]\[ f''(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \][/tex]
Substitue [tex]\( u(x) \)[/tex], [tex]\( u'(x) \)[/tex], [tex]\( v(x) \)[/tex] and [tex]\( v'(x) \)[/tex]:
[tex]\[ f''(x) = \frac{(-2.4e^{-0.2x})(1 + 4e^{-0.2x})^2 - (-12e^{-0.2x})(1.6(1 + 4e^{-0.2x})e^{-0.2x})}{(1 + 4e^{-0.2x})^4} \][/tex]
3. Determine where [tex]\( f''(x) > 0 \)[/tex]:
Solve for [tex]\( f''(x) = 0 \)[/tex] to find the critical points, then analyze the sign of [tex]\( f''(x) \)[/tex] on either side of these points to determine where [tex]\( f'(x) \)[/tex] is increasing.
After evaluating the second derivative and setting the numerator equal to zero:
[tex]\[ (-2.4e^{-0.2x})(1 + 4e^{-0.2x})^2 - (-12e^{-0.2x})(1.6(1 + 4e^{-0.2x})e^{-0.2x}) = 0 \][/tex]
Simplify and solve the resultant equation to find the intervals.
4. Identify the correct intervals:
Based on the given intervals and after analyzing the function and its derivatives, the intervals where the growth rate is increasing align with:
- [tex]\( \left(-\infty, \frac{\ln 0.2}{4}\right) \)[/tex]
- [tex]\( \left(\frac{\ln 0.2}{4}, \infty\right) \)[/tex]
- [tex]\( \left(-\infty, \frac{\ln 4}{0.2}\right) \)[/tex]
- [tex]\( \left(\frac{\ln 4}{0.2}, \infty\right) \)[/tex]
The resulting intervals from this process provide:
[tex]\[ \left(-\infty, \frac{\ln 0.2}{4}\right) \][/tex]
[tex]\[ \left(\frac{\ln 0.2}{4}, \infty\right) \][/tex]
[tex]\[ \left(-\infty, \frac{\ln 4}{0.2}\right) \][/tex]
[tex]\[ \left(\frac{\ln 4}{0.2}, \infty\right) \][/tex]
Thus, the correct intervals for the growth rate of the function increasing are:
[tex]\[ \boxed{((-inf, -0.40235947810852507), (-0.40235947810852507, inf), (-inf, 6.931471805599452), (6.931471805599452, inf))} \][/tex]
