Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

summarize the pertinent information obtained by applying the graphing s…

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 )
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.)
a. 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) ( (-5, infty) ).
(type your answer in interval notation. type an exact answer. use a comma to separate answers as needed.)
a. the function ( f ) is never concave downward.
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the function ( f ) has an inflection point at ( x = )
(type an exact answer. use a comma to separate answers as needed.)
a. the function ( f ) has no inflection point.

Explanation:

Step1: Find the first derivative

The function is \(y = f(x)=\ln(x + 5)-4\).
Using the formula \(\frac{d}{dx}\ln(u)=\frac{u'}{u}\) (where \(u=x + 5\) and \(u'=1\)), we get \(f'(x)=\frac{1}{x + 5}\).

Step2: Find the second derivative

Differentiate \(f'(x)=\frac{1}{x + 5}=(x + 5)^{-1}\) using the power rule \(\frac{d}{dx}x^{n}=nx^{n-1}\).
\(f''(x)=-(x + 5)^{-2}=-\frac{1}{(x + 5)^{2}}\).

Step3: Analyze concavity

For concavity, we consider the sign of \(f''(x)\).
Since \((x + 5)^{2}>0\) for all \(x
eq - 5\) (the domain of \(f(x)\) is \(x>-5\)), then \(f''(x)=-\frac{1}{(x + 5)^{2}}<0\) for all \(x\in(-5,\infty)\).
A function \(y = f(x)\) is concave upward when \(f''(x)>0\) and concave downward when \(f''(x)<0\).
Since \(f''(x)<0\) for all \(x\) in the domain \((-5,\infty)\) of \(f(x)\), there is no interval where \(f''(x)>0\).
An inflection point occurs where \(f''(x) = 0\) or \(f''(x)\) is undefined and the concavity changes. Since \(f''(x)
eq0\) for all \(x\) in the domain of \(f(x)\) (and \(f''(x)\) is defined for all \(x>-5\)), there is no inflection point.

Answer:

  • For concavity upward: B. The function \(f\) is never concave upward.
  • For concavity downward: A. The function \(f\) is concave downward on the sub - interval(s) \((-5,\infty)\)
  • For inflection point: B. The function \(f\) has no inflection point.