(1 point)\nsuppose that on the interval ( i ), ( f(x) ) is positive and concave up. furthermore, assume that…

(1 point)\nsuppose 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.\nto answer the questions, choose your answers from the following list:\ncu (concave up), cd (concave down), ( f(x), f^{prime}(x), f^{prime prime}(x), 0 ), or 1.\na.) ( f^{prime prime}(x)>square ) on ( i )\nb.) ( g^{prime prime}(x)=2left(a^{2}+b f^{prime prime}(x)\right) ), where ( a=square ) and ( b=square )\nc.) ( g^{prime prime}(x)>square ) on ( i )\nd.) ( g(x) ) is ( square ) on ( i )

(1 point)\nsuppose 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.\nto answer the questions, choose your answers from the following list:\ncu (concave up), cd (concave down), ( f(x), f^{prime}(x), f^{prime prime}(x), 0 ), or 1.\na.) ( f^{prime prime}(x)>square ) on ( i )\nb.) ( g^{prime prime}(x)=2left(a^{2}+b f^{prime prime}(x)\right) ), where ( a=square ) and ( b=square )\nc.) ( g^{prime prime}(x)>square ) on ( i )\nd.) ( g(x) ) is ( square ) on ( i )

Answer

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)