Q. 56 - 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.

$$\begin{gathered} f\left(x\right)=5-x^4; \\ [a,b]=[-2,-1]; \\ K=0 \end{gathered}$$

Step-by-Step Solution

Answer

By using the graphing utility, the approximate values of c are $c=-1.5$

Step 1. Given Information.

The function:

$$\begin{gathered} f\left(x\right)=5-x^4; \\ [a,b]=[-2,-1]; \\ K=0 \end{gathered}$$

Step 2. Find the maximum and minimum value.

Substitute the interval values in the given function. 

$$\begin{gathered} f(-2)=5-{(-2)}^4 \\ =5-16 \\ =-11<0 \end{gathered}$$

$$\begin{gathered} f(-1)=5-{(-1)}^4 \\ =5-1 \\ =4>0 \end{gathered}$$

Step 3. Use Intermediate value theorem.

Since f is continuous on $[-2,-1]$ and $f(-2)<0<f(-1)$ by the Intermediate Value Theorem there is some value $c\in (-2,-1)$ for which $f\left(c\right)=0$

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 $f\left(x\right)=5-x^4$ intersects the line $y=0$. From the graph, the value of $f\left(c\right)=0 at c=-1.5$