Wednesday

April 16, 2014

April 16, 2014

Posted by **Kelly** on Tuesday, March 31, 2009 at 5:57pm.

Find g(14).

I know that if (a, f(a)) is on h(x), then (f(a), a) will be on g(x). I also know that if f^-1(a)=b if f(b)=a. I thought that if I could find the x value(s) at which h(x)=14, I would be able to find g(14).

g(x)= x^3+4x-2=14

g(x)= x^3+4x-16=0

I don't know where to go from here. I don't know how to solve the cubic to find x.

My teacher said it was not necessary to find the equation for g(x) to solve the problem. I tried finding it at first, but ran into problems there as well:

if y= x^3+4x-2

x= y^3+4y-2

x-2=y^3+4y

But I don't know how to solve for y.

Any help would be appreciated. Thank you.

- Calculus -
**Count Iblis**, Tuesday, March 31, 2009 at 7:31pmYoiu understand correctly what you need to do, but you should write:

"g(x)= x^3+4x-16=0"

Instead, you can say that if x = g(14), then x satisfies the equation:

x^3+4x-16 = 0

Use the rational roots theorem. Since 16 has many divisors, you can try to shift x, e.g. put x = t + 1.

For the Rational Rpoots theorem, you only need to know the coefficient of t^3 and the constant term. The coefficient of t^3 is 1 and the coefficient of the constant term is the value of the polynomial at t = 0, which corresponds to x = 1, so this is 11.

So, the only possible roots are

t = ±1 and t= ±11.

Add 1 to find the possible roots for the polynomial as a function of x:

x = 0, 2, -10, 12

If we apply the rational roots theorem to the original polynomial directly, then we find that the possible roots are powers of two up to a sign till 16. But we also know that x must be among the above list, so x = 2 is the only possible rational root.

If you try out x = 2, you see that it is indeed a zero.

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