Question

For f(x) = -3x + e^x, show that there is a number c. 0 <= c <= 1, such that f(c) = 0.
Please show steps so I figure out how to do it, thanks!

Answer 1

hint: intermediate value theroem

Answer 2

if you can show there is a value a in [0,1] such that f(a)>0
and a value b in [0,1] such that f(b)<0
then there has to be a value c between a and b such that f(c)=0

Answer 3

and yes you can try pluggin in 0 and 1 to see what happens with the function there.

Answer 4

basically if you can find two numbers a and in the interval [0,1]
such that f(a) and f(b) have different signs then you are done and you can conclude there is some number c in between a and b such that f(c)=0

Answer 5

what is f(0) and f(1) for example?

Answer 6

hmm mkay, sorry this is my first day doing calculus stuff in a while. I got f(0) = 1 and f(1) = -.02817, but I'm still a bit confused as to how to interpret those answers in relation to the rest of the problem.

Answer 7

f(x)=-3x+e^x is continuous function right?

Answer 8

Right! Got that part.

Answer 9

If there are any x-intercepts for f then that means the curve has to go through the x-axis.
That means either before getting to the x-intercept the graph was negative or positive
And after going away from the x-intercept the graph will be the opposite
since the function is either just increasing on 0 to 1 or decreasing on 0 to 1
but you can also verify this by finding f'

Answer 10

so if you want to verify that it is just decreasing or increasing on [0,1]
find f'(x)
and then find the critical number
should be able to tell that critical number is outside the interval which means the function won't change from decreasing to increasing or increasing to decreasing in the interval [0,1]

Answer 11

uh oh, we definitely haven't covered those things yet. We've only reviewed PreCalc and started on the IVT today, so I don't think it will take those into consideration quite yet... hmm. sorry, I'm pretty rusty on my math.

Answer 12

well you really don't need the derivative part

Answer 13

showing f(0) and f(1) are different signs is good enough

Answer 14

when x = 0, y = +1
when x = 1, y = -0.03

y goes from a positive value when x = 0 to a negative value when x = 1.
If it is a continuous curve the ONLY way it can go from a positive y value (that is, above the x-axis) to a negative value (that is, below the x-axis) is if it CROSSES the x-axis somewhere between x = 0 and x = 1. The point where it crosses the x-axis, the y-value will be zero.

Answer 15

you said the function was continuous
i wonder how we connect those dots without going through y=0

Answer 16

we have to go through y=0 don't we?

Answer 17

if we wanted to prove there was exactly 1 zero from [0,1]
we could go further with that f' thing I was talking about
like show that is strictly decreasing on that interval
but we were just asked to show a c exists such f(c)=0

Answer 18

hmm yes exactly, so solving for f(0) & f(1) shows us that it crosses the y axis then...
I GET IT

Answer 19

cool pictures help a lot