Home > Calculus > Analyzing Concavity of a Function

Is #f(x) =e^xsinx-cosx# concave or convex at #x=pi/3#?

Answer 1

Concave up (my naming convention is different to yours, will explain further in explanation)

Obtain #f'(x)# by differentiating the original function (product rule): #f'(x)=e^xsin(x)+e^xcos(x)+sin(x)#

Obtain #f''(x)# by differentiating again: #f''(x)=e^xsin(x)+e^xcos(x)+e^xcos(x)-e^xsin(x)+cos(x)=2e^xcos(x)+cos(x)# #=(2e^x+1)cos(x)#

Evaluate #f''(pi/3)#: #f''(pi/3)=(2e^(pi/3)+1)cos(pi/3)# #=1/2(2e^(pi/3)+1)# Clearly #f''(pi/3)>0#, so the function will be concave up at #x=pi/3#

Note that the naming convention I learnt was concave up and concave down, so I am not sure which is concave or convex in your terms.

Graphing the original fucntion is a good way to check your answer, at #x=pi/3~~1#, we see that the rate of change of gradient is indeed positive and therefore concave up. graph{e^x*sinx-cosx [-3, 3, -5, 5]}

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.