Q. 59 - 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.

$$\begin{gathered} f\left(x\right)=x^3-3x^2-2; \\ [a,b]=[2,4]; \\ K=-4 \end{gathered}$$

Step-by-Step Solution

Answer

By using the graphing utility, the approximate values of c are $c=2.73$

Step 1. Given Information.

The function:

$$\begin{gathered} f\left(x\right)=x^3-3x^2-2; \\ [a,b]=[2,4]; \\ K=-4 \end{gathered}$$

Step 2. Find the maximum and minimum value.

Substitute the interval values in the given function.

$$\begin{gathered} f\left(2\right)=2^3-3{\left(2\right)}^2-2 \\ =8-12-2 \\ =-6<-4 \\ f\left(4\right)=4^3-3{\left(4\right)}^2-2 \\ =64-48-2 \\ =14>-4 \end{gathered}$$

Step 3. Use Intermediate value theorem.

Since f is continuous on $[2,4]$ and $f\left(2\right)<-4<f\left(4\right)$ by the Intermediate Value Theorem

there is some value $c\in (2,4)$ for which $f\left(c\right)=-4$

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 $f\left(x\right)=x^3-3x^2-2$ intersects the line $y=-4$. From the graph, the value of $f\left(c\right)=-4 at c=2.73$