Question

min/max. For what value of X does f(x) take on a minimum if f(x) = 11.4sin(1.7x)-3.4cos(1.7x) within the interval (-1.67, 2.01)? I know I have to take the derivative, then set that equal to 0. Can't find what I'm doing wrong. (my work in reply)

Answer 1

\[f(x) = 11.4\sin(1.7x)-3.4\cos(1.7x)\]
\[f'(x) = 19.38\cos(1.7x)+5.78\sin(1.7x)\]
Now set the derivative to 0 and solve for x:
\[19.38\cos(1.7x)+5.78\sin(1.7x) = 0\]
\[19.38\cos(1.7x)=-5.78\sin(1.7x)\]
\[\frac{ 19.38 }{ -5.78 }=\frac{ \sin(1.7x) }{ \cos(1.7x) }\]
\[\frac{ 19.38 }{ -5.78 }=\tan(1.7x)\]
\[1.7x = \arctan(\frac{ 19.38 }{ -5.78 })\]
\[x = \frac{ 1 }{ 1.7 }(\arctan(\frac{ 19.38 }{ -5.78 })\]
\[x \approx -43.172\]
However, this is not the correct answer. Anyone see what I'm doing wrong?

Answer 2

within the interval (-1.67, 2.01)

Answer 3

So you're not within the interval.

Answer 4

Also, are you in degree mode or radian mode? You're not giving units.

Answer 5

so if tan repeats every pi, between -pi/2 and pi/2, does that mean I have to take that x value and subtract a multiple of pi that brings it within the interval given? say, around 14pi?
yeah, this is all that's given, so I'm guessing whatever is default. degrees?

Answer 6

I think they want radians.
http://www.wolframalpha.com/input/?i=1%2F17+arctan%2819.38%2F-5.78%29

Is this the answer?

Answer 7

I don't have the answer. I have to submit the x value and the min value at the same time and then it'll tell me if they are both right. (I can't check them individually)

Answer 8

Oh, well, I'm pretty sure that -0.07534999 is right

Answer 9

and then plug that into x in the original equation to get the minimum value?

Answer 10

Yeah, you can double check here.

http://www.wolframalpha.com/input/?i=11.4sin%281.7x%29-3.4cos%281.7x%29+minimize

Answer 11

well I entered that the min value occurs at x = -0.07535 and the value of of f(x) at this value of x is -3.42548. (found by plugging in x to f(x)) But it's still saying it's incorrect. Trying to figure out that info from wolfram in terms of the interval.
I'm getting confused. The first link says -0.075, the second says -0.75. This should be a correct query http://www.wolframalpha.com/input/?i=minima+11.4sin%281.7x%29-3.4cos%281.7x%29+over+%28-1.67%2C+201%29 but it gives global values.

Answer 12

ah, I missed a decimal in the input. http://www.wolframalpha.com/input/?i=minima+11.4sin%281.7x%29-3.4cos%281.7x%29+over+%28-1.67%2C+2.01%29 says x= -0.753 and the min is -11.8962. Those values are correct. But I'm still not sure how to get to them and what I needed to do further :(

Answer 13

i think you did all the work

Answer 14

here is the decimal answer to what you wrote above
http://www.wolframalpha.com/input/?i=arctan%28-1938%2F578%29%2F1.7

Answer 15

answer for min should be -11.8962... by a non calculus method

Answer 16

oh, I see. I needed to be in radians mode when solving for x for some reason. Do I need to stay in radians when finding the min? or why did I have to switch to radians?

Answer 17

keep everything in radians when you use calculus, you cannot use degrees

Answer 18

or is radians what the calculator should usually be in and I left it in degree mode on accident last time I used it. I'm confused now after all this.

Answer 19

the derivative of sine is not cosine if you are working in degrees

Answer 20

but there is a simpler method i think

Answer 21

\[a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\sin(x+\theta)\] for suitable \(\theta\)
since the smallest sine can be is \(-1\) your minimum should be
\[-\sqrt{11.4^2+3.4^2}\] which it is
http://www.wolframalpha.com/input/?i=sqrt%2811.4^2%2B3.4^2%29

Answer 22

use degrees if you are "solving" a triangle, otherwise forget about them
sine and cosine are functions of numbers, not angles
and as such they only correspond to the functions of angles if you are measuring your angles in radians
think numbers, not angles

Answer 23

ha, that looks like the method the tutor I first spoke to on campus tried, but he couldn't figure it out.
I think my brain cells that were holding degree/radian info must have died. Thanks for the explanation.
So, I actually worked the problem in a correct way, I just had the calculator in degree mode. Wish I could select two best responses. Thanks for stepping in and clearing some things up. Thanks both of you!

Answer 24

oh wait, so how do I find maximum?

Answer 25

same as the minimum, only with a plus sign for the same reason

Answer 26

largest sine can be is one, so largest
\[\sqrt{a^2+b^2}\sin(x)\] can be is \(\sqrt{a^2+b^2}\)

Answer 27

yeah, I was thinking using the method you showed there with 1 instead would be it. But then how do I first find the x value of the function where the max is using that method?

Answer 28

Also, the theta is tripping me up in that equation. I'm not sure what to plug in there.

Answer 29

i guess you can do what you did before
this problem with all these annoying decimals is a drag

Answer 30

\(\tan(\theta)=\frac{b}{a}\) should do it i think

Answer 31

ok, then even in my method, how do I find the max? I thought to find min/max, you take the derivative of the function, then set it equal to 0. The x values you get (usually you get more than 1) will be used t ofind min/max. Which ever gives greatest y is max, whichever gives smallest y is min. But as shown in my initial work, I only found 1 x value.

Answer 32

you found the min, when you find the arctan you find the number between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\) that works, but tangent is periodic so you can add \(\pi\) so your result, that should give the max i think

Answer 33

no that doesn't work damn

Answer 34

if the values help:
Minimum value of f(x) occurs at x= -0.753499904048986
What is the value of f(x) at this value of x? -11.8962178863704

Maximum value of f(x) occurs at x= 1.09449577453324
What is the value of f(x) at this value of x? 11.8962178863704

Answer 35

ok this work
\[\frac{1}{1.7}\left(\arctan(-\frac{1938}{578})+\pi\right)\]

Answer 36

your min was at
\[\frac{1}{1.7}\arctan(-\frac{1938}{578})\]
you found that one
then you have to add \(\pi\) but do that before dividing

Answer 37

so adding pi will give the max in this situation. So the interval really didn't play a role in this question?

Answer 38

well they were both in the interval
here you can see it
www.wolframalpha.com/input/?i=1%2F1.7+(arctan(19.38%2F-5.78)%2Bpi)

Answer 39

Yeah, they thankfully happened to land in the interval.

Answer 40

not coming up as clickable you can copy and paste and see that it is right
very annoying this question, i hate decimals

Answer 41

if my x values were not in the interval, would I add/subtract multiples of pi until I got a value that was in the interval?

Answer 42

yeah i guess so unless for some reason the interval was so small as not not contain any solution then you would check the endpoints
your function is periodic with period \(\frac{2\pi}{1.7}\) whatever that is

Answer 43

oh, because it was stretched. geez, this question is strange. I hope it's not on an exam considering it took a lot of time and 4-5 people to figure out. I'm good here. Thanks again!