Quiz 3
[Pages:3]Math 0220 Spring 2015
Quiz 3
Solutions
1. (12 pm) For the functions f (x) = cos x and g(x) = 3 7 - x2.
(a) [2 points] Find functions F (x) = f g and G(x) = g f
Solution: F (x) = cos 3 7 - x2
G(x) = 3 7 - cos2 x
(b) [3 points] Find derivatives F (x) and G (x). Do not simplify results.
Solution: F (x) = cos (7 - x2)1/3 = - sin (7 - x2)1/3
1 (7 - x2)-2/3 (-2x) 3
2x sin (7 - x2)1/3
F (x) =
3 (7 - x2)2/3
G (x) = 3 7 - cos2 x
=
(7 - cos2 x)1/3
sin 2x G (x) = 3 (7 - cos2 x)2/3 .
= 1 (7 - cos2 x)-2/3 (-2 cos x)(- sin x) 3
1
1. (1 pm) [5 points] A graph of a function is given. In the coordinate plane below plot the graph of its derivative. Mark all important points.
2
2. (12 pm) [5 points] Find the Slope-Intercept form of the equation of the tangent line to the 2
curve y = at the point where x = 1. x
Find the slope of the tangent line as a limit of difference quotient. No credit will be given if you do not use the limit.
Solution:
22
Slope
m = lim
f (1 + h) - f (1) = lim
- 1+h 1
h0
h
h0
h
2 - 2(1 + h)
= lim h0
1+h h
2 - 2 - 2h = lim
h0 h(1 + h)
= lim
-2h
= lim -2 D=SP
-2 = -2.
h0 h(1 + h) h0 1 + h
1+0
2 y1 = 1 = 2 The tangent line equation: y - 2 = -2(x - 1) y = -2x + 4
2. (1 pm) [5 points] Find an equation of the tangent line to the curve cos(2x + y) = y2 cos x
at the point
4
,
0
.
Write
answer
in
the
Slope-Intercept
form.
Solution: Implicit differentiation: - sin(2x + y)(2 + y ) = 2yy cos x - y2 sin x.
When
x
=
4
and y = 0 we get
- sin
2
(2 + y ) = 0
-2 - y = 0
y
4
= -2 = m.
The tangent line equation:
y - 0 = -2
x
-
4
y
=
-2x
+
2
bonus problem (12 pm) [5 points extra]
f (x) = sin2 x. Find f (9)
4
.
Simplify your answer.
Solution:
f (x) = 2 sin x cos x = sin 2x, f (9)(x) = 28 sin 2x, f (9)
4
= 28 = 256.
(Hint: Use the fact that (sin x)(8) = (sin x)(4) = sin x).
bonus problem (1 pm) [5 points extra]
f (x) = cos2 x. Find f (9)
4
.
Simplify your answer.
Solution:
f (x) = 2 cos x(- sin x) = - sin 2x, f (9)(x) = -28 sin 2x, f (9)
4
= -28 = -256.
(Hint: Use the fact that (sin x)(8) = (sin x)(4) = sin x).
3
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