suppose that ( f(x)=(1 + 7x)e^{3x} ).\n\n a. find all critical values of ( f ). (if there are multiple…

suppose that ( f(x)=(1 + 7x)e^{3x} ).\n\n a. find all critical values of ( f ). (if there are multiple values, enter them separated by commas. if there are no critical points, enter dne.)\n\n ( x=)\n\n b. use interval notation to indicate where ( f ) is increasing. (if there are multiple intervals, enter them separated by commas.)\n\n c. use interval notation to indicate where ( f ) is decreasing. (if there are multiple intervals, enter them separated by commas.)\n\n d. list the ( x )-coordinates of all local maxima of ( f ). (if there are multiple values, enter them separated by commas. if there are no local maxima, enter dne.)
Answer
Explanation:
Step1: Differentiate (f(x)) using product rule
The product rule is ((uv)^\prime = u^\prime v+uv^\prime). Let (u = 1 + 7x) and (v=e^{3x}). Then (u^\prime=7) and (v^\prime = 3e^{3x}). So (f^\prime(x)=7e^{3x}+3(1 + 7x)e^{3x}=(7 + 3+21x)e^{3x}=(10 + 21x)e^{3x}).
Step2: Find critical values
Set (f^\prime(x)=0). Since (e^{3x}>0) for all (x\in R), we solve (10 + 21x=0). Then (x=-\frac{10}{21}).
Step3: Determine intervals of increase and decrease
Test intervals around (x =-\frac{10}{21}).
- For (x<-\frac{10}{21}), let (x=-1). Then (f^\prime(-1)=(10-21)e^{-3}=- 11e^{-3}<0).
- For (x>-\frac{10}{21}), let (x = 0). Then (f^\prime(0)=(10 + 0)e^{0}=10>0).
Answer:
a. (x =-\frac{10}{21}) b. (\left(-\frac{10}{21},\infty\right)) c. (\left(-\infty,-\frac{10}{21}\right)) d. DNE