# For 1990 through 2006, the total exports E (in billions) of the U.S. can be modeled by E = -.13x^3 + 5.033x^2 - 23.2x + 233 where x is the number of years since 1990. During what year were the...

For 1990 through 2006, the total exports E (in billions) of the U.S. can be modeled by E = -.13x^3 + 5.033x^2 - 23.2x + 233 where x is the number of years since 1990. During what year were the total exports about $313 billion?

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### 2 Answers

Given mathematical model for export

`E(x)=-.13x^3+5.033x^2-23.2x+233`

It is given E(x)=313 ,and want to find x ?

i.e.

`313=-.13x^3+5.033x^2-23.2x+233`

`.13x^3-5.033x^2+23.2x+313-233=0`

`.13x^3-5.033x^2+23.2x+80=0`

Sove the cubic equation

x=-2.26,8.31,32.68

-2.26 is not possible because time never negative.

32.68 is nt possible because model is applicable since 1990-2016 ie. max time period=16 years.

**Thus only possible solution is 8.31 years.**

Do you know if you're intended to use a graphing calculator to solve this type of problem? At your level of math, I'm thinking that that's probably what is intended. It is possible to solve in other ways, too, but they're much more difficult.

We know that x represents the number of years since 1990. We're asked to find what year the exports equalled 313 billion. That value must be 1990 + x. So let's find x.

`-.13x^3 + 5.033x^2 - 23.2x + 233 = 313`

`-.13x^3 + 5.033x^2 - 23.2x - 80 = 0`

We want to find where this equation equals zero. That is, the points where the y = f(x) = 0, or the points where our equation/function intersect with the x axis. Using a graphing calculator, you'll find the answers are:

x = -2.26752

x = 8.30504

x = 32.6779