Exercise 59

Stewart Calculus 8e: Section 3.4 - Exercise 59

Page 1 of 2

Exercise 59

Find all points on the graph of the function f (x) = 2 sin x + sin2 x at which the tangent line is horizontal.

Solution

The tangent line to f (x) is horizontal wherever the first derivative is zero. Evaluate the first derivative.

f (x) = df = d (2 sin x + sin2 x) dx dx

d = 2 (sin x) +

d (sin2 x)

dx

dx

d = 2(cos x) + (2 sin x) ? (sin x)

dx

= 2(cos x) + (2 sin x) ? (cos x)

= 2 cos x + 2 sin x cos x

= 2 cos x(1 + sin x)

Set it equal to zero. Solve for x.

2 cos x(1 + sin x) = 0

2 cos x = 0

or 1 + sin x = 0

cos x = 0

or sin x = -1

1

3

x = (2n - 1), n = 0, ?1, ?2, . . . or x = + 2n, n = 0, ?1, ?2, . . .

2

2

Find the corresponding y-values by plugging these values of x into the function for f (x).

f

1 (2n - 1)

= 2 sin

1 (2n - 1)

+ sin2

1 (2n - 1)

= 2(-1)n+1 + 1

2

2

2

f

3 + 2n

= 2 sin

3 + 2n

+ sin2

3 + 2n

= 2(-1) + 1 = -1

2

2

2

Therefore, the points on the curve f (x) = 2 sin x + sin2 x that have a horizontal tangent line are

1 (2n - 1), 2(-1)n+1 + 1

and

3 + 2n, -1 ,

2

2

where n is an integer.



Stewart Calculus 8e: Section 3.4 - Exercise 59 This is confirmed in the graph of f (x) versus x.

Page 2 of 2



 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download