Question

find the intervals on which the graph of ( f ) is concave upward, the intervals on which the graph of ( f ) is concave downward, and the inflection points
( f(x)=24 e^{x}-e^{2 x} )
for what interval(s) of ( x ) is the graph of ( f ) concave upward? select the correct choice below and, if necessary, fill in the answer box to complete your choice.
( (-infty, ln 6) )
(type your answer in interval notation type an exact answer use a comma to separate answers as needed)
the graph is never concave upward.
for what interval(s) of ( x ) is the graph of ( f ) concave downward? select the correct choice below and, if necessary, fill in the answer box to complete your choice
a.
(type your answer in interval notation type an exact answer. use a comma to separate answers as needed)
b. the graph is never concave downward

Explanation:

Step1: Find the first derivative

Using the derivative rules \((e^{ax})^\prime = ae^{ax}\), for \(y = 24e^{x}-e^{2x}\), the first derivative \(y^\prime=f^\prime(x)=24e^{x}-2e^{2x}\).

Step2: Find the second derivative

Differentiate \(y^\prime = 24e^{x}-2e^{2x}\) again. Using the same derivative rule, \(y^{\prime\prime}=f^{\prime\prime}(x)=24e^{x}-4e^{2x}\).

Step3: Set the second derivative equal to zero

Let \(y^{\prime\prime}=0\), so \(24e^{x}-4e^{2x}=0\). Factor out \(4e^{x}\): \(4e^{x}(6 - e^{x})=0\). Since \(e^{x}>0\) for all \(x\in R\), then \(6 - e^{x}=0\), which gives \(e^{x}=6\) and \(x = \ln6\).

Step4: Test intervals for concavity

• For \(x<\ln6\), let \(x = 0\). Then \(y^{\prime\prime}(0)=24e^{0}-4e^{0}=24 - 4=20>0\).
• For \(x>\ln6\), let \(x=\ln(12)\). Then \(y^{\prime\prime}(\ln(12))=24e^{\ln(12)}-4e^{2\ln(12)}=24\times12-4\times144=288 - 576=- 288<0\).

Answer:

• The graph of \(f(x)\) is concave upward on the interval \((-\infty,\ln6)\).
• The graph of \(f(x)\) is concave downward on the interval \((\ln6,\infty)\).