1. ## Intermediate Value Theorem

Use the Intermediate Value Theorem to show that the polynomial has a zero in the given interval.

f(x) = 3x^4 + 4x^3 - 8x - 2; [1, 2]

Also, I noticed that there is no x^2 term. Does this change the process in terms of solving the question?

2. Originally Posted by magentarita
Use the Intermediate Value Theorem to show that the polynomial has a zero in the given interval.

f(x) = 3x^4 + 4x^3 - 8x - 2; [1, 2]

Also, I noticed that there is no x^2 term. Does this change the process in terms of solving the question?

Let $\displaystyle f(x)=3x^4+4x^3-8x-2$. Since f is a polynomial, it is continuous on $\displaystyle [1, 2]$.

We have $\displaystyle f(1)=3(1)^4+4(1)^3-8x-2=-3 < 0 \ \ and \ \ f(2)=3(2)^4+4(2)^3-8x-2=62 > 0$, so that $\displaystyle f(1)<0<f(2)$.

So by the intermediate-value theorem there exists $\displaystyle x_1$ in $\displaystyle (1, 2)$ such that $\displaystyle f(x_1)=0$.

That is, the equation $\displaystyle 3x^4+4x^3-8x-2=0$ has a solution in the interval $\displaystyle (1, 2)$.

3. ## ok....

Originally Posted by masters
Let $\displaystyle f(x)=3x^4+4x^3-8x-2$. Since f is a polynomial, it is continuous on $\displaystyle [1, 2]$.

We have $\displaystyle f(1)=3(1)^4+4(1)^3-8x-2=-3 < 0 \ \ and \ \ f(2)=3(2)^4+4(2)^3-8x-2=62 > 0$, so that $\displaystyle f(1)<0<f(2)$.

So by the intermediate-value theorem there exists $\displaystyle x_1$ in $\displaystyle (1, 2)$ such that $\displaystyle f(x_1)=0$.

That is, the equation $\displaystyle 3x^4+4x^3-8x-2=0$ has a solution in the interval $\displaystyle (1, 2)$.
Thank you very much.