Question

does f(x) = x^3 - 1 + sin(x) have a root in (0,1)? Explain

Answer 1

no??

Answer 2

ahhhh

Answer 3

thank-you so much! im suppose to justify each step. doesn't feel like there's much to justify.

Answer 4

basically says: Explain your answer
carefully, and be sure to justify each assertion you make.

Answer 5

seems so unnecessary

Answer 6

haha, I agree. Do you have time to help me with 1 more q?

Answer 7

let a and be denote constants. find the values of a and b so that the function (1+e^-x if x<0 and ax+b if x>or=0) is differentiable everywhere

Answer 8

"and b"

Answer 9

im following

Answer 10

when i solve for a, I will also have x

Answer 11

how will i isolate a

Answer 12

x could be equal to greater than 0?!
so I could set x as 1, when solving for a?

Answer 13

i am so wrong.

Answer 14

ugh

Answer 15

b=2; a=0

Answer 16

since the function is differentiable everywhere,
f' (x) at x=0 exist
so, f'(x) at x->0- = f'(x) at x->0+
d/dx (1+e^-x) = d/dx (ax+b) at x=0

Answer 17

you can get a from here

Answer 18

tell me if you did not get what i did

Answer 19

gotchya

Answer 20

so what you get for a ?

Answer 21

o :S

Answer 22

derivative of 1+e^-x
is
-e^-x right ?
when x=o, left side = -e^0 = -1

got this ?

Answer 23

yes, got that

Answer 24

right side ?

Answer 25

derivative of ax+b is just a
so, overall, a = -1 :)

Answer 26

eeeeek, how come it's -?

Answer 27

b=2; a=-1 then?

Answer 28

comeee back, hartnnnn.

Answer 29

yes, thats correct.

Answer 30

you got the steps, right ? and how it is - ?

Answer 31

hartnn answered your second question

the idea is that the function
1+e^(-x) --> 2 as x-->0, i.e. the point (0,2)
you want the second function, defined for x≥0 to start at (0,2) so that you do not have a discontinuity.

in other words, you want a x + b= 2 when x is 0. that gives you b=2

you also want the function to be differentiable. That means you want the same differential for 1+e^(-x) as x-->0 as you get for a x + b at x=0

the differential of 1+e^(-x) = -e^(-x) (take the derivative)
as x->0, -e^(-x) --> -1
one way to see this:
\[ -e^{-x}= \frac{-1}{e^x} \\
\lim_{x->0} \frac{-1}{e^x}= \frac{-1}{1}= -1
\]

that means you want the derivative of a x + b to match -1 when x is 0
d (ax+b)= a
you want a= -1

the final answer is -x+2

here is a graph of the situation

Answer 32

For your first question
***does f(x) = x^3 - 1 + sin(x) have a root in (0,1)? Explain***

the interval does not include 0 or 1 (the parens mean exclude the boundary values). However, if we evaluate f(x) at the boundaries we find
f(0)= 0 - 1 + sin(0)= -1
f(1)= 1-1 +sin(1) = 0.84 (x is assumed to be in radians)

by the intermediate value theorem
http://www.mathsisfun.com/algebra/intermediate-value-theorem.html

there must be some x within the closed interval [0+, 1-]
where f(x)=0
(by [0+,1-] I mean the interval +episilon to 1 - epsilon, because I think the intermediate value theorem wants a closed interval.