MATH 147
Quiz Key #4
2/25/2016
(1) Find an equation for the tangent line to the curve y = 7x3 + 2x − 1 at
x = 1. Express your answer in slope-intercept form.
Solution: Let f ( x ) = 7x3 + 2x − 1. Then we need to find the slope of the
tangent line and a point on the tangent line. Since
f (1) = 7 + 2 − 1 = 8,
a point on the tangent line is given by (1, 8) . The slope of the tangent line is
given by f 0 (1) .
f 0 ( x ) = 21x2 + 2.
Therefore, f 0 (1) = 21 + 2 = 23. Hence, an equation in point-slope form for the
tangent line is given by
y − 8 = 23 ( x − 1) .
y − 8 = 23x − 23.
y = 23x − 15.
(2) Use the Product Rule to differentiate f ( x ) =
1 2
2x
−1
2x + 3x2 .
Solution: Note that
1
i
d 1 2
d h
2
0
2
2x + 3x2 .
f (x) =
x − 1 2x + 3x +
x −1
dx 2
2
dx
1 2
∴ f 0 ( x ) = x 2x + 3x2 +
x − 1 (2 + 6x ) .
2
∴ f 0 ( x ) = 3x3 + 2x2 + 3x3 + x2 − 6x − 2.
∴ f 0 ( x ) = 6x3 + 3x2 − 6x − 2.
1
(3) Use the Quotient Rule to differentiate h (t) =
t2 − 3t + 1
.
t+1
Solution:
0
h (t) =
d
dt
t2 − 3t + 1 (t + 1) − t2 − 3t + 1
( t + 1)
0
h (t) =
2
(2t − 3) (t + 1) − t2 − 3t + 1
( t + 1)2
∴ h0 (t) =
2t2 − t − 3 − t2 + 3t − 1
∴ h0 (t) =
d
dt
( t + 1)2
t2 + 2t − 4
( t + 1)2
[ t + 1]
.
.
.
.
2