Question

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: Recall concavity condition

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

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

We know that \(g(x)=(f(x))^{2}\).
Using the chain rule, the first derivative \(g'(x)=2f(x)f'(x)\).
Using the product rule \((uv)^\prime = u^\prime v+uv^\prime\) where \(u = 2f(x)\) and \(v = f'(x)\), the second derivative \(g''(x)=2(f'(x))^{2}+2f(x)f''(x)\).
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 \((f'(x))^{2}\geq0\). Then \(g''(x)=2((f'(x))^{2}+f(x)f''(x))>0\) on \(I\) (because \(f(x)>0\) and \(f''(x)>0\) and \((f'(x))^{2}\geq0\)).

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

Since \(g''(x)>0\) on \(I\), by the second - derivative test for 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. \(CU\)