# Find the intervals on which the function is concave up or down, the points of inflection, and the...

## Question:

Find the intervals on which the function is concave up or down, the points of inflection, and the critical points, and determine whether each critical point corresponds to a local minimum or maximum (or neither). Let

{eq}f(x) = 5x + 5\sin (x),0 \le x \le 2\pi {/eq}

## Extremes:

To find the local extremes we can follow the following steps:

1. Find the first derivative and calculate its critical points.

2. Apply a criterion of the first derivative:

- If the first derivative is positive to the left of the point and negative to the right, then it is a maximum.

- If the first derivative is negative to the left of the point and positive to the right, then it is a minimum.

- If the first derivative does not change sign around the critical point, this point is not a local maximum or minimum of the function.

Monotony and extremes

Find the critical point

{eq}\displaystyle f(x) = 5x + 5\sin (x)\\ \displaystyle f'(x) = 5 + 5\cos (x)\\ \displaystyle f'(x)=0 \,\,\,\, \rightarrow \,\,\,\, 5 + 5\cos (x)=0\\ \displaystyle 5\cos (x)=-5\\ \displaystyle \cos (x)=-1\\ \displaystyle x=\pi +2k\pi \ \ \forall k \in Z\\ \boxed{\displaystyle x=\pi} \Longrightarrow \textrm {critical point of the function on the given interval} {/eq}

Then, to find the open intervals on which the function is increasing or decreasing, we look within our range: {eq}0 \le x \le 2\pi{/eq} and separate that range by our critical point, {eq}\pi {/eq} , creating intervals where the first derivative can possibly change sign changes sign which are {eq}\left (0, \pi \right) \,\,\,\,\, \left (\pi,2\pi \right) {/eq}

To define the sign of the first derivative in each interval, we evaluate a point of each interval and verify the sign:

{eq}\left (0, \pi \right) \,\,\, \rightarrow \,\,\, f'(\frac {\pi}{2})=5 > 0 \,\,\, \rightarrow \,\,\,\, \textrm {the first derivative is positive in this interval}\\ \left (\pi,2\pi \right) \,\,\, \rightarrow \,\,\, f'(\frac {3\pi}{2})=5 > 0 \,\,\, \rightarrow \,\,\,\, \textrm {the first derivative is positive in this interval} {/eq}

Conclusion (Monotony)

{eq}\boxed{ \left (0, \pi \right) \bigcup \left (\pi,2\pi \right)} \,\, \Longrightarrow \,\, \textrm {the function is increasing}\\ {/eq}

Therefore,

{eq}\boxed{\textrm {The function has no local extremes points on the given interval}}\\ {/eq}

Concavity and points of inflection

To determine the function's concavity, we find and analyze the second derivative (point of inflection and concavity):

{eq}\displaystyle f'(x) = 5 + 5\cos (x)\\ \displaystyle f''(x) = -5\sin (x)\\ \displaystyle f''(x) = 0 \,\, \rightarrow \,\, x= k\pi \, k \, \in \, Z\\\\ \displaystyle x= \pi \,\, \rightarrow \,\, \textrm {point where the second derivative is zero} {/eq}

To define the sign of the second derivative in each interval, evaluate a point of each interval and verify the sign:

{eq}\left (0, \pi \right) \,\,\, \rightarrow \,\,\, f''(\frac {\pi}{2})=-5 < 0 \,\,\, \rightarrow \,\,\,\, \textrm {the second derivative is negative in this interval}\\ \left (\pi,2\pi \right) \,\,\, \rightarrow \,\,\, f''(\frac {3\pi}{2})=5 > 0 \,\,\, \rightarrow \,\,\,\, \textrm {the second derivative is positive in this interval} {/eq}

Conclusion (Concavity and points of inflection) {eq}\displaystyle f(\pi) = 5\pi\\ \boxed{ \left (\pi,5\pi \right)} \textrm {Inflection point of the function}\\ \boxed{ \left (0, \pi \right) } \,\, \Longrightarrow \,\, \textrm {concave down}\\ \boxed{ \left (\pi,2\pi \right) } \,\, \Longrightarrow \,\, \textrm {concave up}\\ {/eq}