Respuesta :
Answer:
The maximum value of the function is 0.168 at x=1.5.
Step-by-step explanation:
The given function is
[tex]f(x)=x^3e^{-2x}[/tex]
We need to find the maximal value of this function.
Differentiate the given function with respect to x.
[tex]f'(x)=\frac{d}{dx}x^3e^{-2x}[/tex]
Product rule of differentiation:
[tex]\frac{d}{dx}(fg)=fg'+f'g[/tex]
Using product rule, we get
[tex]f'(x)=x^3\frac{d}{dx}e^{-2x}+e^{-2x}\frac{d}{dx}x^3[/tex]
[tex]f'(x)=x^3e^{-2x}(-2)+e^{-2x}(3x^2)[/tex]
[tex]f'(x)=x^2e^{-2x}(-2x+3)[/tex] .... (1)
Equate f'(x)=0.
[tex]0=x^2e^{-2x}(-2x+3)[/tex]
[tex]x^2=0\Rightarrow x=0[/tex]
[tex]e^{-2x}=0\Rightarrow \text{no solution}[/tex]
[tex]-2x+3=0 \Rightarrow x=1.5[/tex]
The function has two critical values 0 and 1.5.
Differentiate the f'(x) with respect to x.
[tex]f''(x)=2 e^{-2 x} x (3 - 6 x + 2 x^2)[/tex]
At x=0,
[tex]f''(0)=2 e^{-2 (0)} (0)(3 - 6 (0) + 2*(0)^2)=0[/tex]
Since f''(0)=0, therefore x=0 is the point of inflection.
At x=1.5,
[tex]f''(1.5)=2 e^{-2 (1.5)} (1.5)(3 - 6 (1.5) + 2*(1.5)^2)\approx-0.224[/tex]
Since f''(1.5)<0, therefore the function is maximum at x=1.5.
Substitute x=1.5 in the given equation.
[tex]f(1.5)=(1.5)^3e^{-2(1.5)}[/tex]
[tex]f(1.5)\approx 0.168[/tex]
Therefore the maximum value of the function is 0.168 at x=1.5.
