consider the equation below. (if an answer does not exist, enter dne.)\n\n$f(x)=e^{9x}+e^{-x}$\n\n(a) find…

consider the equation below. (if an answer does not exist, enter dne.)\n\n$f(x)=e^{9x}+e^{-x}$\n\n(a) find the interval on which $f$ is increasing. (enter your answer using interval notation.)\n\nfind the interval on which $f$ is decreasing. (enter your answer using interval notation.)\n\n(b) find the local minimum and maximum values of $f$.\n\nlocal minimum value\n\nlocal maximum value\n\n(c) find the inflection point.\n\n$(x,y)=( )$\n\nfind the interval on which $f$ is concave up. (enter your answer using interval notation.)\n\nfind the interval on which $f$ is concave down. (enter your answer using interval notation.)

consider the equation below. (if an answer does not exist, enter dne.)\n\n$f(x)=e^{9x}+e^{-x}$\n\n(a) find the interval on which $f$ is increasing. (enter your answer using interval notation.)\n\nfind the interval on which $f$ is decreasing. (enter your answer using interval notation.)\n\n(b) find the local minimum and maximum values of $f$.\n\nlocal minimum value\n\nlocal maximum value\n\n(c) find the inflection point.\n\n$(x,y)=( )$\n\nfind the interval on which $f$ is concave up. (enter your answer using interval notation.)\n\nfind the interval on which $f$ is concave down. (enter your answer using interval notation.)

Answer

Explanation:

Step1: Find the first derivative

Using the chain rule, if (y = e^{u}), then (y^\prime=e^{u}\cdot u^\prime). For (f(x)=e^{9x}+e^{-x}), (f^\prime(x)=9e^{9x}-e^{-x}=\frac{9e^{10x} - 1}{e^{x}})

Step2: Find critical points

Set (f^\prime(x) = 0), so (9e^{10x}-1 = 0). (e^{10x}=\frac{1}{9}), then (10x=\ln(\frac{1}{9})=-\ln(9)), (x =-\frac{\ln(9)}{10}=-\frac{\ln(3^{2})}{10}=-\frac{\ln(3)}{5})

Step3: Determine increasing and decreasing intervals

  • Test an interval to the left of (x =-\frac{\ln(3)}{5}), say (x=-1). (f^\prime(-1)=9e^{-9}-e^{1}<0)
  • Test an interval to the right of (x =-\frac{\ln(3)}{5}), say (x = 0). (f^\prime(0)=9 - 1=8>0)

So (f(x)) is decreasing on ((-\infty,-\frac{\ln(3)}{5})) and increasing on ((-\frac{\ln(3)}{5},\infty))

Step4: Find local minimum and maximum

Since (f(x)) changes from decreasing to increasing at (x =-\frac{\ln(3)}{5}), the local minimum value is (f(-\frac{\ln(3)}{5})=e^{9(-\frac{\ln(3)}{5})}+e^{-(-\frac{\ln(3)}{5})}=e^{-\frac{9\ln(3)}{5}}+e^{\frac{\ln(3)}{5}}=\frac{1}{3^{\frac{9}{5}}}+3^{\frac{1}{5}}=\frac{1 + 3^{2}}{3^{\frac{9}{5}}}=\frac{10}{3^{\frac{9}{5}}}) and there is no local maximum (because the function only has one critical point and the function changes from decreasing to increasing)

Step5: Find the second derivative

(f^\prime(x)=9e^{9x}-e^{-x}), then (f^{\prime\prime}(x)=81e^{9x}+e^{-x}=\frac{81e^{10x}+1}{e^{x}}) Since (e^{10x}>0) for all (x\in R), (f^{\prime\prime}(x)>0) for all (x\in R)

Answer:

(a) Increasing: ((-\frac{\ln(3)}{5},\infty)), Decreasing: ((-\infty,-\frac{\ln(3)}{5})) (b) Local minimum value: (\frac{10}{3^{\frac{9}{5}}}), Local maximum value: DNE (c) Inflection point: DNE, Concave up: ((-\infty,\infty)), Concave down: DNE