Question

Help please!!! (:
Identify the inflection points and local max and min of the function below.

Answer 1

\[f(x)=2sinx-x\] on the interval \[[\frac{ - \pi }{ 2 } , \frac{ \pi }{ 2 } ]\]

Answer 2

Okay; first, observe that local extrema occur (for continuous functions) where our function is locally 'flat' like the peak of a hill, so that if you move immediately in either direction you won't immediately be at a more extreme value. To state this more precisely, we use the *derivative* to tell us the local behavior of a continuous function; at a local extremum, the derivative is \(0\) (this is equivalent to what I said above).

Answer 3

Differentiating the function we get \(f'(x)=2\cos x-1\) and we solve for where it vanishes to determine *critical points* (candidates for local extrema):$$f'(x)=0\\2\cos x-1=0\\2\cos x=1\\\cos x=\frac12$$On the interval \([-\pi/2,\pi/2]\), what solutions do we have to the above?

Answer 4

1/2?

we find f"(x) to find the critical points right?

Answer 5

@oldrin.bataku

Answer 6

your solution is above, only thing left to do is solve
\[\cos(x)=\frac{1}{2}\] for \(x\)

Answer 7

\[x=\frac{ 1/2 }{ \cos }\] ? @satellite73

Answer 8

@pdd21, \(\cos x\) is a function, not a coefficient. You should have \(x=\cos^{-1}\left(\dfrac{1}{2}\right)=\cdots\)

Answer 9

so not that I solved f'(x) I got f"(x)=-2sinx and that would equal \[\sin ^{-1} (\frac{ 1 }{ -2 })\]

Answer 10

@sithsandgiggles

Answer 11

Yes, you would have an inflection point for when \(x=\sin^{-1}\left(-\dfrac{1}{2}\right)\).

Answer 12

how would I figure out the local extrema of the function on the given interval and where they occur? @SithsAndGiggles

Answer 13

hold the phone!

Answer 14

what you need is to figure out what number (or angle) \(x\) has \(\cos(x)=\frac{1}{2}\)

Answer 15

In general, let's say you're given a function defined for all \(x\) in \(-\infty<x<\infty\).

Now let's say you find \(n\) critical points to be \(x=x_1,x_2,\cdots,x_n\). You would then split up the domain interval \(\left((-\infty,\infty)\right)\) into \(n+1\) intervals:
\[(-\infty,x_1),~(x_1,x_2),~\cdots,~(x_n,\infty)\]
Next, you pick a value of \(x\) within each interval and plug it into the derivative function. In other words, find \(f'(\overline{x})\) for some \(\overline{x}\) in any given interval. If \(f'(\overline{x})<0\), that means \(f(x)\) is decreasing for that particular interval. If \(f'(\overline{x})>0\), that means \(f(x)\) is increasing for the interval. (If it's zero, that means there's no change.)

You determine whether a maximum or minimum occurs at each of the critical points based on how you find out the function behaves on each interval in between critical points.

An increase-decrease indicates a maximum.
A decrease-increase indicates a minimum.
An increase-increase or decrease-decrease likely indicates you made a mistake somewhere along the line.

Answer 16

saying
\[\cos(x)=\frac{1}{2}\] is solved via
\[x=\frac{\frac{1}{2}}{\cos}\] is like saying
\[f(x)=\frac{1}{2}\] means
\[x=\frac{\frac{1}{2}}{f}\]

Answer 17

go to the unit circle and find an angle where \(\cos(x)=\frac{1}{2}\) that is all you are looking for

Answer 18

cos 60?

Answer 19

@satellite73

Answer 20

so x=60 @satellite73

Answer 21

I'd suggest sticking with radians, and not degrees. \(\cos60^\circ=\cos\dfrac{\pi}{3}\)

However, keep in mind your interval: \(f(x)\) is defined for \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\). There are two values of \(x\) in this interval that satisfy \(\cos x=\dfrac{1}{2}\).

Answer 22

so how do I go about doing that... I'm really confused.. @sithsandgiggles