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On what interval(s) is the function f(x) = x^3e^(2x) increasing?
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guess you have to take the derivative and figure out where it is positive (the derivative that is). did you get the derivative?
\[f'(x)=3x^2e^{2x}+2x^3e^{2x}\] \[=e^{2x}(3x^2+2x^3)\]
1) (-inf, -3/2], [0,inf) 2) (-inf, -3/2] 3) (-inf, 0], [3/2, inf) 4) (-inf, 3/2] 5) [-3/2, inf)
since \[e^{2x}>0\] for all x, you job is to look at \[3x^2+2x^3=x^2(3+2x)\]
and since \[x^2\geq0\] you only need to solve \[3+2x>0\] \[3>2x\] \[\frac{3}{2}>x\]
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looks like answer 4
nopee :/
yeah because my algebra stinks it is \[3+2x>0\] \[3>-2x\] \[-\frac{3}{2}>x\]
lol. i still dont know which interval ;/
maye 1 or 2? idk..
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