# Problems getting started on this problem

1. Sep 17, 2006

### donjt81

Hi I am trying to solve the following problem.

f(x) = 2x^3 - 5x - 5
show that there is atleast one value of c for which f(x) = pie

I am assuming they want me to use pie = 3.14

so I did the following

2x^3 - 5x - 5 = 3.14

but then I dont know how to solve for x can someone please give me some kinda hint to get me started?

Thanks

2. Sep 17, 2006

### shmoe

I sure hope they don't!

f(x) = 2x^3 - 5x - 5 is a continuous function (right?), what kinds of theorems or results do you have on continuous functions that might be useful here?

3. Sep 17, 2006

### donjt81

ok so i have to use the intermediate value theorem. i still dont understand the pie thing... if its not 3.14 then what do they want me to use for pie.

so with the IVT i would do the following

1] define an inteval [a, b] such that when i find f(a) and f(b) one value is bigger than "pie" and the other value is less than "pie"

2] I would also have to show that f is continous in that interval [a, b]

so for example if i find out that f(a) < "pie" < f(b) then i can say that yes there does exist a point c where f(c) = "pie"

but I dont know what "pie" is i guess so I'm not sure what I am looking for.

4. Sep 17, 2006

### shmoe

pi=3.141526535... it does not equal any truncated sum with a finite number of digits (it also has no "e" on the end). For your purposes you can probably safely assume some bounds like 3<=pi<=4 and find an a and b where:

f(a) < 3 <= pi <= 4 < f(b)

as for continuity, f(x) is a polynomial. You should know something about polynomials and continuity.

5. Sep 18, 2006

### donjt81

thanks got it