Q. 63 Use the Intermediate Value Theor... [FREE SOLUTION]

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Chapter 1: Q. 63 (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.
$f (x) = \sin x; K = \frac{1}{2}$

Short Answer

Expert verified

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

Step by step solution

Step 1. Given Information.

The function:

$f (x) = \sin x; K = \frac{1}{2}$

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 $\frac{1}{2}$ .

We know that,

$\sin 0 = 0 < \frac{1}{2} \sin \frac{蟺}{4} = \frac{1}{\sqrt{2}} > \frac{1}{2}$

Step 3. Apply intermediate value theorem.

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

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

From this graph we can conclude that $f (c) = \frac{1}{2} a t c = \frac{1}{2}$

$f (c) = \frac{1}{2} a t c = \frac{1}{2}$

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Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Approximate the interval.

Step 3. Apply intermediate value theorem.

Step 4. Approximate c.

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