Step by step solution to "Is f(x)=exsin⁡x−cos⁡x concave or convex at x=π3?"

Raven Gardner

Answered question

2022-04-13

f (x) = e^{x} \sin x - \cos x

concave or convex at

x = \frac{π}{3}

Answer & Explanation

Frain4i62

Beginner2022-04-14Added 16 answers

Step 1
Obtain f'(x) by differentiating the original function (product rule):

f^{'} (x) = e^{x} \sin (x) + e^{x} \cos (x) + \sin (x)

Obtain f"(x) by differentiating again:

f^{″} (x) = e^{x} \sin (x) + e^{x} \cos (x) + e^{x} \cos (x) - e^{x} \sin (x) + \cos (x) = 2 e^{x} \cos (x) + \cos (x)

= (2 e^{x} + 1) \cos (x)

Step 2
Evaluate

f \frac{π}{3})

f \frac{π}{3}) = (2 e^{\frac{π}{3}} + 1) \cos (\frac{π}{3})

= \frac{1}{2} (2 e^{\frac{π}{3}} + 1)

Clearly

f \frac{π}{3}) > 0

, so the function will be concave up at

x = \frac{π}{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 = \frac{π}{3} \approx 1

, we see that the rate of change of gradient is indeed positive and therefore concave up.

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