OpenStudy (anonymous):

Can someone assist me in this question... I understand the premise, but i'm confused on how to approach it... "prove that the equation has at least one real root. 100e^-x/100=.01x^2

2 years ago
OpenStudy (kainui):

What do you know about the intermediate value theorem?

2 years ago
OpenStudy (anonymous):

I know that you set the equation equal to zero, then you choose an interval for two x values and plug them into the equation to get the y-values, and if you have a positive and negative value that proves that there is a root of the function as long as it is continuous, but i think where im confused is how would I set this equal to zero?.. and how do i know that it's continuous.?

2 years ago
OpenStudy (anonymous):

without using a graphing calculator.

2 years ago
OpenStudy (kainui):

What makes you suspicious that this isn't continuous?

2 years ago
OpenStudy (anonymous):

well because of e^negative value is going to be all positive values?... so i'm not sure theyre would be any portion of the graph that would crossover the x-axis?

2 years ago
OpenStudy (kainui):

You're completely right, it will be completely positive. =) So when you said earlier " set the equation equal to zero" What do you mean by this?

2 years ago
OpenStudy (anonymous):

i guess what i'm trying to say is without knowing what this graph really looks like and trying to find a real "root", how would i approach it so i'm not trying to look through every single x-value when it may not even have a root?... i'm confused..

2 years ago
OpenStudy (kainui):

Here we're not really looking for roots, we're really looking for when the graph f(x)=100e^-x/100 intersects the graph g(x)=.01x^2 and if the question really says " has at least one real root." I can see why that would be confusing, since you're really just trying to show that there is some value of x where these two functions are equal.

2 years ago
OpenStudy (kainui):

I'll be back in about 30 minutes if you still need help, I have to go eat dinner. =P

2 years ago
OpenStudy (anonymous):

o.k. thank you

2 years ago