Question

(1 point)
suppose that on the interval ( i ), ( f(x) ) is positive and concave up. furthermore, assume that ( f^{prime prime}(x) ) exists and let ( g(x)=(f(x))^{2} ). use this information to answer the following questions.
to answer the questions, choose your answers from the following list:
cu (concave up), cd (concave down), ( f(x), f^{prime}(x), f^{prime prime}(x), 0 ), or 1.
a.) ( f^{prime prime}(x)>square ) on ( i )
b.) ( g^{prime prime}(x)=2left(a^{2}+b f^{prime prime}(x)
ight) ), where ( a=square ) and ( b=square )
c.) ( g^{prime prime}(x)>square ) on ( i )
d.) ( g(x) ) is ( square ) on ( i )

Explanation:

Step1: Use the concavity property

If a function \(y = f(x)\) is concave - up on an interval \(I\), then by the definition of concavity, \(f''(x)>0\) on \(I\).

Step2: Find the first and second derivatives of \(g(x)\)

We know that \(g(x)=(f(x))^{2}\).
Using the chain rule, \(g'(x) = 2f(x)f'(x)\).
Using the product rule \((uv)^\prime=u^\prime v + uv^\prime\) where \(u = 2f(x)\) and \(v=f'(x)\), we have:
\(g''(x)=2(f'(x))^{2}+2f(x)f''(x)\).
Comparing with \(g''(x)=2(A^{2}+Bf''(x))\), we get \(A = f'(x)\) and \(B=f(x)\).

Step3: Analyze the sign of \(g''(x)\)

Since \(f(x)>0\) and \(f''(x)>0\) on \(I\), and \((f'(x))^{2}\geq0\) for all \(x\) (because the square of a real number is non - negative).
\(g''(x)=2((f'(x))^{2}+f(x)f''(x))\).
Since \((f'(x))^{2}\geq0\), \(f(x)>0\), and \(f''(x)>0\), then \(g''(x)>0\) on \(I\).

Step4: Determine the concavity of \(g(x)\)

If \(g''(x)>0\) on an interval \(I\), then by the definition of concavity, the function \(y = g(x)\) is concave - up (\(CU\)) on \(I\).

Answer:

a) \(0\)
b) \(A = f'(x)\), \(B = f(x)\)
c) \(0\)
d) \(CU\)