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.

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 upward.

Explanation:

Step1: Find the first derivative

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

Step2: Find the second derivative

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

Step3: Set the second derivative greater than 0 to find concave - upward intervals

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 \(x\in R\), we solve \(6 - e^{x}>0\).
\(e^{x}<6\). Taking the natural logarithm of both sides (\(y = \ln x\) is an increasing function), we get \(x<\ln6\).

Answer:

A. \((-\infty,\ln6)\)