Resolved: "Is f(x)=sin⁡x concave or convex at x=π5?"

Briley Cabrera

Answered question

2022-04-15

f (x) = \sin x

concave or convex at

x = \frac{π}{5}

entreblogsmc2j

Beginner2022-04-16Added 10 answers

Step 1
f(x) is concave at

x_{0} if f (x_{0}) > \frac{1}{2} (f (x_{0} - h) + f (x_{0} + h))

and
f(x) is convex at

x_{0}

f (x_{0}) < \frac{1}{2} (f (x_{0} - h) + f (x_{0} + h))

.
where

(x_{0} - h)

and

(x_{0} + h)

are two values around

x_{0}

.
As

\frac{π}{5} = 36^{\circ}

, let us calculate sin x at

{x = 35^{\circ}, 36^{\circ}, 37^{\circ}}

{\sin 35}^{\circ} = 0.5735764

{\sin 36}^{\circ} = 0.5877853

{\sin 37}^{\circ} = 0.6018150

Step 2
Now as

\frac{1}{2} (\sin^{\oplus} {\sin 37}^{\circ}) = \frac{1}{2} (0.5735764 + 0.6018150)

= \frac{1}{2} \times 1.1753914 = 0.5876957 < {\sin 36}^{\circ}

f (x) = \sin x

is concave at

\frac{π}{5}

Further if f(x) can be differentiated twice at

x = x_{0}

and

f x_{0}) < 0

, it is concave and if

f x_{0}) > 0

, it is convex.
As

f (x) = \sin x, f^{'} (x) = \cos x and f x) = - \sin x

and

f 36^{\circ}) = - {\sin 36}^{\circ} = - 0.5877853

and hence

f (x) = \sin x

is concave at

\frac{π}{5}

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?