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 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.
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the function ( f ) is concave downward 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 downward.

Explanation:

Step1: Find the first derivative

The derivative of \(y = f(x)=\ln(x + 5)-4\) using the formula \(\frac{d}{dx}\ln(u)=\frac{u'}{u}\) (where \(u=x + 5\) and \(u'=1\)). So \(f'(x)=\frac{1}{x+5}\).
Set \(f'(x) = 0\), \(\frac{1}{x+5}=0\) has no solution. Since the domain of \(y=\ln(x + 5)-4\) is \(x>-5\) and \(f'(x)=\frac{1}{x + 5}>0\) for all \(x>-5\), the function is increasing on \((-5,\infty)\). A local minimum occurs where the function changes from decreasing to increasing. Since the function is always increasing, there is no local minimum.

Step2: Find the second derivative

Differentiate \(f'(x)=\frac{1}{x+5}=(x + 5)^{-1}\) using the power rule \(\frac{d}{dx}u^n=nu^{n - 1}u'\). Here \(n=-1\) and \(u=x + 5\), \(u'=1\). So \(f''(x)=-(x + 5)^{-2}=-\frac{1}{(x + 5)^2}\).
Since \(f''(x)=-\frac{1}{(x + 5)^2}<0\) for all \(x>-5\) (because \((x + 5)^2>0\) for \(x>-5\)), the function is concave - downward on \((-5,\infty)\) and never concave - upward.

Answer:

B. The function \(f\) has no local minimum.
B. The function \(f\) is never concave upward.
A. The function \(f\) is concave downward on the sub - interval \((-5,\infty)\)