Question

f(x) = x^3*cos(x) - x*sin(x) + 1

in the interval [-2,2]

they want me to prove that there are at least two zeros within the interval... what does that mean? like, where they cross the axis? do you plug in zeros for the x? how do you find more than one zero? there's only one solution if you plug in the zero?

Answer 1

if it really says prove there are at least two zeros, then since this funciton is continuous see if you can find two changes in sign, from postitive to negative and back say
if it changes sign twice, that means it must cross the x - axis twice
try easy number like
\[\frac{\pi}{2}, -\frac{\pi}{2}\] etc

Answer 2

what do you mean by try? like, substitue for x?

Answer 3

sorry, i'm REALLY behind

this website was my last ditch effort to trty to understand... I need a tutor BAD

Answer 4

yes i mean check with numbers

Answer 5

ok..
according to intermediate value theorem on the interval [a, b]. If d [f (a), f (b)], then there is a c [a, b] such that f (c) = d.
you should first find f(0) and f'(0) if you compute them, you will see f(0)= -1 and f'(0)=0 which are both in the interval

Answer 6

for example, at x = 0 you get 1 right?

Answer 7

\[f(0) = 0^3\cos(0) - 0\sin(0) + 1=1 \]

Answer 8

so we know when x = 0 is it positive. now try at
\[x=-\frac{\pi}{2}\]and i bet you will get a negative number

Answer 9

f'(x)= 3x^2*cosx-x^3*sinx-sinx-x.cosx

Answer 10

f'(0)= 0 and f(0)=1 both are in the interval so there are at least two zeros

Answer 11

if fact i am sure of it, you will get
\[f(-\frac{\pi}{2})=-\frac{\pi}{2}+1\] which is negative

Answer 12

similarly
\[f(\frac{\pi}{2})=-\frac{\pi}{2}+1\] also negative. so it goes from negative to positive and then back to negative. therefore it must have at least two zeros

Answer 13

awesome. thank guys!