BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 1, Problem 18RE
To determine

To find: An expression function for the graph which satisfies the given conditions.

Expert Solution

Answer to Problem 18RE

Solution:

The equation of the function is f(x)={2x2 on [2,1)1x2 on [1,1].

Explanation of Solution

Given:

The graph has a line segment connecting (−2, 2) and (−1, 0) and it consist a top half of the circle with centre (0, 0) and radius 1.

Calculation:

Find the slope of the line segment joining the points (−2, 2) and (−1, 0) as follows.

m=y2y1x2x1

m=021(2)=21=2

Thus, the slope of the line segment is m=2.

Find the y-intercept of the line segment joining the points (−2, 2) and (−1, 0) as follows.

y=mx+c0=(2)(1)+c[m=2]0=2+cc=2

Thus, y-intercept is c=2.

Therefore, the equation of the line segment is, y=2x2 where 2x<1 (1)

Also there exist top half of the circle with centre (0, 0) and radius 1.

The standard equation of the circle which passes through the (0, 0) and radius 1 is x2+y2=1.

Solve the equation for y as follows.

y2=1x2y=±1x2

Since the graph consist only top half of the circle, consider only the positive root. That is y=1x2.

Therefore, equation of the top half of the circle is y=1x2,1x1 (2)

Combine the equations (1) and (2), the function of the graph is f(x)={2x2 on [2,1)1x2 on [1,1].

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