summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of ( y =…

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of ( y = f(x) ).\n( f(x)=ln (x + 5)-4 )\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. the function ( f ) is concave upward on the subinterval(s)\n(type your answer in interval notation. type an exact answer. use a comma to separate answers as needed.)\na. the function ( f ) is never concave upward.\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. the function ( f ) is concave downward on the subinterval(s) ( (-5, infty) ).\n(type your answer in interval notation. type an exact answer. use a comma to separate answers as needed.)\na. the function ( f ) is never concave downward.\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. the function ( f ) has an inflection point at ( x = )\n(type an exact answer. use a comma to separate answers as needed.)\na. the function ( f ) has no inflection point.
Answer
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\neq - 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)\neq0) 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.