Question

Finding x intercepts of f(x)= 6sinx-x^2?

Answer 1

I can get to x^2=6sinx but then how do i reduce it down?

Answer 2

Can you use a calculator?

Answer 3

One obvious solution is \(x = 0\), but I don't see how you'll get the other ones (if there are any other ones).

Answer 4

... without using a calculator, I mean.

Answer 5

sorry!

Answer 6

Gah, had to leave computer. i think my teacher wants this long hand, so no calculator.

Answer 7

you have to find the x and y intercepts. wouldnt that just give you the y ones?

Answer 8

@electrokid

Answer 9

x-intercept = when function crosses x-axis
when it does, y=0. solve for x

Answer 10

right, so 0=6sinx-x^2

Answer 11

yep.. what are the solutions?

Answer 12

thats the part im unsure of, what do i do with sinx?

Answer 13

how about x=0?

Answer 14

its 0

Answer 15

so, x=0 is one of the x-intercept

Answer 16

when else can this happen?

Answer 17

No idea. (0,0) is a point though so thats something

Answer 18

Dont you do that when you're finding the concavity?

Answer 19

did you study the numerical methods like Bisection method, Newton-Raphson, Secant method?

Answer 20

uhhh no.

Answer 21

without that, analytically, I do not see a way to get the second root.

Answer 22

or replace sine function by exponentials

Answer 23

Well, If you can show id be grateful. its possible my teacher forgot to mention.

Answer 24

\[0={6\over2i}(e^{ix}-e^{-ix})-2x^2\]

Answer 25

you mean the interpolation?

Answer 26

using the numerical method?

Answer 27

yes. to find the umm intercepts

Answer 28

ok, let us say the second root is approximately at "x=3"
you correct it by the following:
\[\color{red}{\Large\boxed{x_{\rm new}=x_{\rm old}-\frac{f(x)}{f'(x)}}}\]

Answer 29

\[x_{\rm new}=3-\frac{6\sin(3)-2(3^2)}{6\cos(3)-4(3)}\approx2.044\]

Answer 30

How would you get 3?

Answer 31

oh wait.. i took the wrong function...
about the "3", ypu can start with any approximation,.. may be "4".. its fine.. you'd slowly reach the correct value, number of steps will change..

Answer 32

ugh. this sounds complicated. Is this calculator version any easier?

Answer 33

would you recommend it?

Answer 34

\[x_{\rm new}=3-\frac{6\sin(3)-(3^2)}{6\cos(3)-2(3)}\approx2.317\\
x_{\rm new}=2.317-\frac{6\sin(2.317)-(2.317)^2}{6\cos(2.317)-2(2.317)}\approx\boxed{2.21}\]

Answer 35

the second accurate intercept is 2.2014

Answer 36

This is called Newton-Raphson interpolation scheme.. if you are taking Calc1, you may aswell understand it and use it!

Answer 37

haha alright. how do you achieve the 2.317?

Answer 38

use a calculator! :)

Answer 39

out of curiosity...
check if they meant \[f(x)=6\sin(x-x^2)\]

Answer 40

well 2.2 is exactly what my calculator graphs it as. there is a space between the x and sin...no parenthesis though.

Answer 41

I assume if it was 6sin(x-x^2) it would have to have them?

Answer 42

if it was the way I just said, I'd be very simple to evaluate..
\[0=6\sin(x-x^2)\\
x-x^2={n\pi}\qquad\text{n=0,+-1, +-2, +-3,...}\]

Answer 43

that would make more sense considering i just did a \[2\sqrt(x)-x \] problem

Answer 44

Facepalm

Answer 45

thank you

Answer 46

quadratic. yep.

Answer 47

you are welcome.