Answer:
[tex]f(\frac{-1}{b})=\frac{1}{e}[/tex]
Step-by-step explanation:
1) According to Fermat Theorem, local extrema can only occur at critical points. So let's evaluate the 1st derivative of that function to find a critical point, that in case of f'(c)=0 a critical point may be a local extrema.
[tex]y=axe^{bx}\\\frac{\mathrm{d} }{\mathrm{d} x}[axe^{bx}]\Rightarrow y'=a(xe^{bx})'+(a)'xe^{bx}\Rightarrow y'=\left(abx+a\right)\mathrm{e}^{bx}\\\left(bx+1\right)\mathrm{ae}^{bx}=0\Rightarrow x=\frac{-1}{b}\\[/tex]
2) Plugging in the point [tex]x=\frac{-1}{b}[/tex] in the function, for nonzero a and b to find the absolute minimum value of that function:
[tex]f(x)=axe^{bx}\Rightarrow f(\frac{-1}{b})=a(\frac{-1}{b})e^{b\frac{-1}{b}}\Rightarrow f(\frac{-1}{b})=a(\frac{-1}{b})e^{-1}\Rightarrow f(\frac{-1}{b})=-e^{-1}\therefore f(\frac{-1}{b})=\frac{1}{e}[/tex]
