Question

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of ( y = f(x) ).
( f(x)=ln (x + 5)-4 )
a. the function ( f ) has a local maximum at ( x = )
(type an exact answer. use a comma to separate answers as needed.)
b. the function ( f ) has no local maximum.
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the function ( f ) has a local minimum at ( x = )
(type an exact answer. use a comma to separate answers as needed.)
b. the function ( f ) has no local minimum.
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the function ( f ) is concave upward on the subinterval(s)
(type your answer in interval notation. type an exact answer. use a comma to separate answers as needed)
b. the function ( f ) is never concave upward.

Explanation:

Step1: Find the domain

For \(y = \ln(x + 5)-4\), the argument of the logarithm must be positive. So \(x+5>0\), which gives \(x>- 5\). The domain is \((-5,\infty)\).

Step2: Find the first - derivative

Using the chain rule, if \(y=\ln(u)-4\) where \(u = x + 5\), then \(y^\prime=\frac{dy}{du}\cdot\frac{du}{dx}\). Since \(\frac{d}{du}(\ln(u))=\frac{1}{u}\) and \(\frac{d}{dx}(x + 5)=1\), we have \(y^\prime=\frac{1}{x + 5}\).
Set \(y^\prime = 0\), \(\frac{1}{x+5}=0\) has no solution. So there are no critical points for local maxima or minima.

Step3: Find the second - derivative

Differentiate \(y^\prime=\frac{1}{x + 5}=(x + 5)^{-1}\) using the power rule \((u^n)^\prime=nu^{n - 1}u^\prime\). Here \(n=-1\) and \(u=x + 5\), \(u^\prime = 1\). So \(y^{\prime\prime}=-(x + 5)^{-2}=-\frac{1}{(x + 5)^2}\).
Since \(y^{\prime\prime}<0\) for all \(x\in(-5,\infty)\) (because \((x + 5)^2>0\) for \(x>-5\)), the function is concave down on \((-5,\infty)\) and never concave up.

Answer:

For local maximum: B. The function \(f\) has no local maximum.
For local minimum: B. The function \(f\) has no local minimum.
For concavity: B. The function \(f\) is never concave upward.