CindyHoth412
99

# F(x)=(x-6)e^-3x find the interval in which f(x) is increasing, the intervals on which f(x) is decreasing, and the local extrema

$\bf f(x)=(x-6)e^{-3x}\\\\ -----------------------------\\\\ \cfrac{dy}{dx}=1\cdot e^{-3x}+(x-6)-3e^{-3x}\implies \cfrac{dy}{dx}=e^{-3x}[1-3(x-6)] \\\\\\ \cfrac{dy}{dx}=e^{-3x}(19-3x)\implies \cfrac{dy}{dx}=\cfrac{19-3x}{e^{3x}}$ set the derivative to 0, solve for "x" to get any critical points keep in mind, setting the denominator to 0, also gives us critical points, however, in this case, the denominator will never be 0, so... no critical points from there there's only 1 critical point anyway, and do a first-derivative test on it, check a number before it and after it, to see what sign the derivative has, and thus, whether the graph is going up or down, to check for any extrema