Q. 63 - 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.

$$\begin{gathered} f\left(x\right)=\sin x; \\ K=\frac{1}{2} \end{gathered}$$

Step-by-Step Solution

Answer

The approximate value of c of the function at $K=\frac{1}{2}$ is $c=\frac{1}{2}$.

Step 1. Given Information.

The function: 

$$\begin{gathered} f\left(x\right)=\sin x; \\ K=\frac{1}{2} \end{gathered}$$

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 $\frac{1}{2}$.

We know that,

 $$\begin{gathered} \sin 0=0<\frac{1}{2} \\ \sin \frac{\mathrm{\pi }}{4}=\frac{1}{\sqrt{2}}>\frac{1}{2} \end{gathered}$$

Step 3. Apply intermediate value theorem.

Since f is continuous on $[0,\frac{\mathrm{\pi }}{4}]$ and , by the Intermediate Value Theorem there is some value $c\in [0,\frac{\mathrm{\pi }}{4}]$ for which $f\left(c\right)=\frac{1}{2}$.

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 $f\left(c\right)=\frac{1}{2}$ by approximating the values
of x for which the graph of $f\left(x\right)=\sin x$ intersects the line $y=\frac{1}{2}$.

From this graph we can conclude that $f\left(c\right)=\frac{1}{2} at c=\frac{1}{2}$

$f\left(c\right)=\frac{1}{2} at c=\frac{1}{2}$