Q. 64 Use the Intermediate Value Theor... [FREE SOLUTION]

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Chapter 1: Q. 64 (page 121)

Use the Intermediate Value Theorem to show that for each function fand value K in Exercises 61鈥�66, there must be some 肠鈭圧 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.
$f (x) = \sin x; K = \frac{\sqrt{3}}{2}$

Short Answer

Expert verified

The approximate value of c of the function at $K = \frac{\sqrt{3}}{2}$ is $c 鈮� 1.05$ .

Step by step solution

Step 1. Given Information.

The function:

$f (x) = \sin x; K = \frac{\sqrt{3}}{2}$

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 $K = \frac{\sqrt{3}}{2}$ .

$f (\frac{蟺}{4}) = \sin \frac{蟺}{4} = \frac{1}{\sqrt{2}} < \frac{\sqrt{3}}{2} f (\frac{蟺}{2}) = \sin \frac{蟺}{2} = 1 > \frac{\sqrt{3}}{2}$

Step 3. Apply intermediate value theorem.

Since f is continuous on $[\frac{蟺}{4}, \frac{蟺}{2}]$ and $f (\frac{蟺}{4}) < \frac{\sqrt{3}}{2} < f (\frac{蟺}{2})$ , by the Intermediate Value Theorem there is some value $c$ for which $f (c) = \frac{\sqrt{3}}{2}$ .

Note that the Intermediate Value Theorem doesn鈥檛 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 $f (c) = \frac{\sqrt{3}}{2}$ by approximating the values
of x for which the graph of $f (x) = \sin x$ intersects the line $y = \frac{\sqrt{3}}{2}$ .