Intermediate Value Theorem

Prove that $x^3+x-4=0$ has at least one solution. Then find the solution.

Step 1: Define f(x) and graph

We'll let $f\left(x\right)=x^3+x-4$

Step 2: Define a y-value for c

From the graph and the equation, we can see that the function value at $c$ is 0.

Step 3: Ensure f(x) meets the requirements of the IVT

From the graph and with a knowledge of the nature of polynomial functions, we can confidently say that $f\left(x\right)$ is continuous on any interval we choose.

We can see that the root of $f\left(x\right)$ lies between 1 and 1.5. So, we'll let our interval be [1, 1.5]. The Intermediate Value Theorem says that $f\left(c\right)=0$ must lie between $f\left(a\right)$ and $f\left(b\right)$. So, we plug in and evaluate $f\left(1\right)$ and $f(1.5)$.

$\begin{array}{rcl}f\left(1\right)& <& f\left(c\right)<f(1.5)\\ 1^3+1-4& <& 0<1.5^3+1.5-4\\ -2& <& 0<0.875\end{array}$

Step 4: Apply the IVT

Now that all of the IVT requirements are met, we can conclude that there is a value $c$ in $[1,1.5]$ such that $f\left(c\right)=0$.

So, $f\left(x\right)$ is solvable.

Does the function $f\left(x\right)=x^2$ take on the value $f\left(x\right)=7$ on the interval $[1,4]$?

Step 1: Ensure f(x) is continuous

Next, we check to make sure the function fits the requirements of the Intermediate Value Theorem.

We know that $f\left(x\right)$ is continuous over the entire interval because it is a polynomial function.

Step 2: Find the function value at the endpoints of the interval

Plugging in $x=1$ and $x=4$ to $f\left(x\right)$

$$\begin{gathered} f\left(1\right)=1^2=1 \\ f\left(4\right)=4^2={16}^{} \end{gathered}$$

Step 3: Apply the Intermediate Value Theorem

Obviously, $1<7<16$. So we can apply the IVT.

Now that all IVT requirements are met, we can conclude that there is a value $c$ in [1, 4] such that $f\left(c\right)=7$.

Thus, $f\left(x\right)$ must take on the value 7 at least once somewhere in the interval $[1,4]$.

Remember, the IVT guarantees at least one solution. However, there may be more than one!

Prove the equation $\frac{x-1}{x^2+2}=\frac{3-x}{1+x}$ has at least one solution on the interval $[-1,3]$.

Let's try this one without using a graph.

Step 1: Define f(x)

To define $f\left(x\right)$, we'll factor the initial equation.

$\begin{array}{rcl}(x-1)(x+1)& =& (3-x)(x^2+2)\\ x^2-1& =& -x^3+3x^2-2x+6\\ x^3-2x^2+2x-7& =& 0\end{array}$

So, we'll let $f\left(x\right)=x^3-2x^2+2x-7$

Step 2: Define a y-value for c

From our definition of $f\left(x\right)$ in step 1, $f\left(c\right)=0$.

Step 3: Ensure f(x) meets the requirements of the IVT

From our knowledge of polynomial functions, we know that $f\left(x\right)$ is continuous everywhere.

We will test our interval bounds, making $a=-1andb=3$. Remember, using the IVT, we need to confirm

$$\begin{gathered} f\left(a\right)<f\left(c\right)<f\left(b\right) \\ f\left(a\right)<0<f\left(b\right) \end{gathered}$$

Let $a=-1$:

$\begin{array}{rcl}f\left(a\right)& =& f(-1)\\ & =& {(-1)}^3-2{\left(-1\right)}^2+2\left(-1\right)-7\\ & =& -12\end{array}$

$\begin{array}{rcl}f\left(b\right)& =& f\left(3\right)\\ & =& 3^3-2{\left(3\right)}^2+2\left(3\right)-7\\ & =& 8\end{array}$

Therefore, we have

$$\begin{gathered} f\left(a\right)<f\left(c\right)<f\left(b\right) \\ -12<0<8 \end{gathered}$$

Therefore, but the IVT, we can guarantee there is at least one solution to

$x^3-2x^2+2x-7=0$

on the interval $[-1,3]$.

Step 4: Apply the IVT

Now that all IVT requirements are met, we can conclude that there is a value $c$ in [0, 3] such that $f\left(c\right)=0$.

So, $f\left(x\right)$ is solvable.