find the derivative of f(x) = (x - 3)e^x

[Ashley5]

find the derivative of f(x) = (x - 3)e^x

I don't know where to begin, I've never seen this problem before. Please help. Thank you!

 

[jwpaine]

Use product rule:

f(x)=(x-3)e^x
f'(x)=d/dx(x-3)e^x + (x-3)d/dx(e^x)

Now take the derivative of the terms according to the product rule.

remember d/dx(e^x) = e^x
because with logarithmic differentiation (showing chain rule):

for f(x) = e^x
ln(f(x)) = ln(e^x)
ln(f(x)) = xln(e)
ln'(f(x)) = d/dx(x)
f(x)(1/f(x))f'(x) = 1f(x) <--- chain rule on LHS
f'(x) = f(x)
f'(x) = e^x

 

[Ashley5]

so, the answer would be:
f'(x)= e^x+ (x-3)(e^x)

but how do you find the increasing and decreasing and any local extrema? I think I took the the derviative wrong

 

[skeeter]

so, the answer would be:
f'(x)= e^x+ (x-3)(e^x)

but how do you find the increasing and decreasing and any local extrema?

find the x-values where f'(x) = 0 and determine if f'(x) changes sign at those x-values ...

e^x + (x-3)e^x = 0

factor out e^x from both terms ...

e^x[1 + (x-3)] = 0

combine like terms ...

e^x(x - 2) = 0

since e^x is always positive, x = 2 is your only critical value where f'(x) = 0.

does f'(x) change sign at x = 2? if so, does it change from (+) to (-) indicating a max, or from (-) to (+) indicating a min ?

 

[Ashley5]

yeah, I got it! Thank you so much!