Question

Anyone good with calculus able to help find an inflection point. f'(x)=x/sqrt(2)+cos(x)

Answer 1

(a) First find critical points (points where f ' = 0)

f(x) = (cosx)^2 - 2sinx on [0, 2pi]

f ' (x) = 2cosx * (-sinx) - 2cosx
f ' (x) = 0 = -2sinxcosx - 2cosx
-2cosx (sinx + 1) = 0
-2cosx = 0 or sinx+1=0
cosx = 0 or sinx = -1
x = pi/2, 3pi/2 or x = 3pi/2
x = pi/2 or 3pi/2

Now we must check when f ' is positive or negative on the follow intervals:

[0, pi/2] and [pi/2, 3pi/2] and [3pi/2, 2pi]

To check the sign we pick a point in each interval the sign of f ' at that point tells us the sign of f ' throughout that interval.

Start with [0, pi/2]. Pick x = pi/4

f ' (pi/4) = -2(sqrt(2)/2)(sqrt(2)/2) - 2(sqrt(2))/2 = -1-sqrt(2) < 0. Thus, f ' < 0 on [0, pi/2]

Next is [pi/2, 3pi/2]. Pick x = pi

f ' (pi) = -2*(0)*(-1) - 2*(-1) = 2 > 0. Thus, f ' > 0 on [pi/2, 3pi/2]

Finally is [3pi/2, 2pi]. Check x = 7pi/4

f ' (7pi/4) = -2*(-sqrt(2)/2)*(sqrt(2)/2) - 2(sqrt(2)/2) = 1-sqrt(2) < 0. Thus, f ' < 0 on [3pi/2, 2pi] also.

Remember: If f ' > 0 (f ' < 0) on [a,b], then f is increasing (decreasing) on [a,b].

Thus, we have found (a) and (b)

(a) f is increasing on [pi/2, 3pi/2] and f is decreasing on [0, pi/2] U [3pi/2, 2pi]

(b) Local minimum at x = pi/2 and x = 2pi; Local maximum at x = 0 and x = 3pi/2

Now apply these steps to f '' to find (c).

We know f ' (x) = -2sinxcosx - 2cosx. Then

f '' (x) = -2(cosx)^2 + 2(sinx)^2 + 2sinx
0 = -2(cosx)^2 + 2(sinx)^2 + 2sinx

Remember (cosx)^2 = 1 - (sinx)^2

0 = -2(1 - (sinx)^2) + 2(sinx)^2 + 2sinx
0 = -2 + 2(sinx)^2 + 2(sinx)^2 + 2sinx
0 = 2(2(sinx)^2 + sinx - 1)
0 = 2(sinx)^2 + sinx - 1

Let y = sinx for a moment. Then

0 = 2y^2 + y - 1
0 = 2y^2 + (2y - y) - 1 since y = 2y-y
0 = (2y^2+2y) + (-y-1)
0 = (2y)(y+1) + -1(y+1)
0 = (2y-1)(y+1)
2y-1 = 0 or y+1 = 0
y = 1/2 or y = -1

Now replace y with sinx

sinx = 1/2 or sin x = -1
x = pi/6 or 5pi/6 or x = 3pi/2
x = pi/6, 5pi/6, 3pi/2

Check the signs like before on the following intervals:

[0, pi/6)
(pi/6, 5pi/6)
(5pi/6, 3pi/2)
(3pi/2, 2pi]

Answer 2

@vexus Where did the f(x) equation you are using come from?.

Answer 3

I know how to do the whole, let f'(x)=0 to find the critical points. but just need to work out where to go from. \[-x/\sqrt{2}=\cos(x)\]

Answer 4

\[f'(x)=\frac{x}{\sqrt2}+\cos x~~\Rightarrow~~f''(x)=\frac{1}{\sqrt2}-\sin x\]
Set \(f''(x)=0\) and solve for the critical points:
\[\begin{align*}
\frac{1}{\sqrt2}-\sin x&=0\\
\sin x&=\frac{1}{\sqrt2}\\
x&=\frac{\pi}{4}+2n\pi,~\frac{3\pi}{4}+2n\pi
\end{align*}\]

Answer 5

ahh thanks, didn't realise that you could work out the inflection point by looking at the f''(x). I am assuming this wouldn't work for max and min since they just require +ve or -ve, not a set number? @SithsAndGiggles

Answer 6

To clarify, \(f''(x)\) helps you determine the inflection points for \(f(x)\), not \(f'(x)\). That's what I assumed, that you're given the derivative for some "unknown" \(f(x)\), and you are to find the inflection points for \(f(x)\).

As for finding extrema: you won't be able to use the second derivative to find the extrema of \(f(x)\); that's what the first derivative is for.

You *can* however find the extrema of the derivative function by using the second derivative. Is this what you mean?

Answer 7

yeah, I was giving the function f(x)=\[f(x)=\frac{ x^2 }{ 2*2^.5}+\sin(x)\]
and has to find if there were critical points between (0,pi/2) and to identify the inflection. and got stuck after differentiating

Answer 8

Okay, so in this context the critical points are \(x\) values for which \(f'(x)=0\), and the inflection points are \(x\) for which \(f''(x)=0\).

The work for the critical points is likely beyond manual computation, so you'll have to refer to a graphing calculator or computer or something to solve for them. If you're expected to only say whether or not an extremum occurs at all, you can plug in various values of \(x\) from the \(\left(0,\dfrac{\pi}{2}\right)\) into the derivative and checking whether the derivative changes from positive to negative.

The inflection points are easier to find. In fact, I've already done that for you :). There's only one, at \(x=\dfrac{\pi}{4}\).

Answer 9

A Warm Welcome To Open Study

Answer 10

Ok was just bugging me that I couldn't work out the critical. Exam in 2 days, thats one less thing to worry about. So thank you very much

Answer 11

You're welcome!

Answer 12

Wouldn't have any Idea about finding the points on a circle that form 2 tangent lines that intercept outside of the circle itself?

Answer 13

If it'll take a while don't worry about it, don't want to take up to much of your time.

Answer 14

What is the point that the tangent lines are touching? It looks like the origin, but can't be sure

Answer 15

oh it gives you the origin, it ask as what points on the circle, creates those 2 tangents, I have tried a few different ways but could never get it.
the equation is x^2-4x+y^2+3=0 ..... (x-2)^2+y^2=1. which is that circle with radius of 1 and a center as (2,0)

Answer 16

it is in the implicit differention section, so I have been using that. but It is very annoying, tried simultaneous using y-y1=m(x-x1) and changing findings the points of the tangent to values of x. since both x's are the same, but it get a value that is out of the circle.

Answer 17

The tangent line to any point on the circle has slope \(\dfrac{dy}{dx}\):
\[\begin{align*}(x-2)^2+y^2&=1\\
2(x-2)+2y\frac{dy}{dx}&=0\\
2x-4+2y\frac{dy}{dx}&=0\\
\frac{dy}{dx}&=\frac{4-2x}{2y}\\
\frac{dy}{dx}&=\frac{2-x}{y}
\end{align*}\]
Looking at a graph, you should expect one tangent line per semicircle, as in your drawing.
\[\text{upper semicircle: }y=\sqrt{1-(x-2)^2}\\
\text{lower semicircle: }y=-\sqrt{1-(x-2)^2}\]
So for a tangent line on the upper semicircle, you have
\[\frac{dy}{dx}=\frac{2-x}{\sqrt{1-(x-2)^2}}\]
and similarly for the lower semicircle,
\[\frac{dy}{dx}=\frac{x-2}{\sqrt{1-(x-2)^2}}\]

Answer 18

Focus on the upper semicircle, and suppose the tangent line touches the circle at some point \((a,b)\).

The slope of this line would be
\[\frac{dy}{dx}=\frac{2-a}{\sqrt{1-(a-2)^2}}\]
And knowing that this line passes through the origin, you set up the equation:
\[\begin{align*}y-0&=\frac{2-a}{\sqrt{1-(a-2)^2}}(x-0)\\
y&=\frac{2-a}{\sqrt{1-(a-2)^2}}x\end{align*}\]
For the lower semicircle, there's only a slight difference. Graphically, you can see that the corresponding point on the circle would look like \((a,-b)\). The slope would be
\[\frac{dy}{dx}=\frac{a-2}{\sqrt{1-(a-2)^2}}\]
and so the lower tangent line would be
\[y=\frac{a-2}{\sqrt{1-(a-2)^2}}x\]

Answer 19

From here, you have to find the actual values of \(a\) and \(b\).

The tangent line for the upper half circle and the upper half circle itself intersect at \((a,b)\), so you get the equation
\[\underbrace{\frac{2-a}{\sqrt{1-(a-2)^2}}x}_{\text{tangent line}}=\underbrace{\sqrt{1-(x-2)^2}}_{\text{upper semicircle}}\]
Solving for \(x\) should give the \(x\)-coordinate of the point on the circle.

Answer 20

Sorry, I overlooked something. Plug in \(x=a\), then you have
\[\underbrace{\frac{2-a}{\sqrt{1-(a-2)^2}}a}_{\text{tangent line}}=\underbrace{\sqrt{1-(a-2)^2}}_{\text{upper semicircle}}\]
Now solve for \(a\).

Answer 21

I solve a = 5/6, or and i number, but 5/6 doesn't fit in the circle.

Answer 22

\[\begin{align*}\frac{2-a}{\sqrt{1-(a-2)^2}}a&=\sqrt{1-(a-2)^2}\\
a(2-a)&=1-(a-2)^2\\
2a-a^2&=1-a^2+4a-4\\
2a&=3\\
a&=\frac{3}{2}
\end{align*}\]

Answer 23

yeah that was me being dumb, I had the 4a and ive and the 4 as positive.

Answer 24

You should be able to find \(b\) pretty easily.

Here's what I'm getting for the tangent lines:
http://www.wolframalpha.com/input/?i=Plot%5B%7BSqrt%5B1-%28x-2%29%5E2%5D%2C-Sqrt%5B1-%28x-2%29%5E2%5D%2C1%2FSqrt%5B3%5D*x%2C-1%2FSqrt%5B3%5D*x%7D%2C%7Bx%2C0%2C3%7D%5D

Answer 25

Well thank you very much, When I first attempted my brain didnt click that it would be (a,b)and (a,-b) so I had it down as (a,b) (a,c) which screwed me over

Answer 26

That process would have worked, you would still find that \(c=-b\).

Answer 27

but because of that I I didn't see you could just put 0,0 in and was putting y-b=m(x-a) and y-c=m(x-a) and then getting annoyed I couldn't do that.. well Thank you so much for all of the help

Answer 28

though inb4 it doesn't even mention it on the exam

Answer 29

No problem