Q. 60 - Calculus

Question

Use the Intermediate Value Theorem to show that for each function f , interval [a, b], and value K in Exercises 55– 60, there is some c∈(a, b) for which f(c) = K. Then use a graphing utility to approximate all such values c. You may assume that these functions are continuous everywhere.

$$\begin{gathered} f\left(x\right)=2+x+x^3; \\ [a,b]=[-1,2]; \\ K=3 \end{gathered}$$

Step-by-Step Solution

Answer

By using the graphing utility, the approximate values of c are $c=0.68$

Step 1. Given Information.

The function:

$$\begin{gathered} f\left(x\right)=2+x+x^3; \\ [a,b]=[-1,2]; \\ K=3 \end{gathered}$$

Step 2. Find the maximum and minimum value.

Substitute the interval values in the given function. 

$$\begin{gathered} f(-1)=2+(-1)+{(-1)}^3 \\ =2-1-1 \\ =0<3 \\ f\left(2\right)=2+2+2^3 \\ =12>3 \end{gathered}$$

Step 3. Use Intermediate value theorem.

Since f is continuous on $[-1,2]$ and $f(-1)<3<f\left(2\right)$ by the Intermediate Value Theorem there is some value $c\in (-1,2)$ for which $f\left(c\right)=3$

Step 4. Sketch the function.

Graph the function in the given interval. 

Step 5. Find c.

From the graph, we can approximate the values of x for which the function $f\left(x\right)=2+x+x^3$ intersects the line $y=3$. From the graph, the value of $f\left(c\right)=3 at c=0.68$