Exercise 59
Stewart Calculus 8e: Section 3.4 - Exercise 59
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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.
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