Q. 65 - Calculus

Question

Use the Intermediate Value Theorem to show that for each function f and value K in Exercises 61–66, there must be some c∈R for which f(c) = K. You will have to select an appropriate interval [a, b] to work with. Then find or approximate one such value of c. You may assume that these functions are continuous everywhere.

$$\begin{gathered} f\left(x\right)=\left|3x+1\right|; \\ K=1 \end{gathered}$$

Step-by-Step Solution

Answer

The approximate value of c of the function at $K=1$ is $c=1$.

Step 1. Given Information.

The function: 

$$\begin{gathered} f\left(x\right)=\left|3x+1\right|; \\ K=1 \end{gathered}$$

Step 2. Approximate the interval.

By trial and error we can find such values a and b, by testing different values of f(x) until we find one that is less than and one that is greater than $1$.

$$\begin{gathered} f\left(0\right)=\left|3\left(0\right)+1\right| \\ =\left|1\right|=1 \\ f\left(1\right)=\left|3\left(1\right)+1\right| \\ =4>1 \end{gathered}$$

Step 3. Apply intermediate value theorem.

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

Note that the Intermediate Value Theorem doesn’t tell us where c is, only that such a c exists somewhere in the interval.

Step 4. Approximate c.

We can approximate some values of c for which $f\left(c\right)=1$ by approximating the values
of x for which the graph of $f\left(x\right)=\left|3x+1\right|$ intersects the line $y=1$.

From this graph we can conclude that $f\left(c\right)=1 at c\approx 1$