Download 1. Find the function y = f(x) and input value a for which the

1.
Find the function y = f (x) and input value a for which the expression
lim
ex
x→0
2
−1
− 1/e
x
represents f 0 (a).
Solution: We know that
f 0 (a) = lim
h→0
f (x) − f (a)
f (a + h) − f (a)
= lim
.
x→a
h
x−a
In this case we use the second expression defining f 0 (a), so we should have
f (x) − f (a)
ex
f (a) = lim
. = lim
x→a
x→0
x−a
0
so we conclude f (x) = ex
2.
2
−1
2
−1
− 1/e
,
x
, a = 0.
Express as a limit the slope of the tangent line of f (x) = ln
√
x2 − 3 at (2, f (2)).
Solution: Since the slope of the tangent line to y = f (x) at (a, f (a)) is given by f 0 (a), using that f (2) = 0
we conclude
f 0 (2) = lim
h→0
3.
ln
p
(2 + h)2 − 3
h
.
Using the definition of f 0 (a) (by limits), find the equation of the tangent line to f (x) = −x3 at (−1, f (−1)).
Solution: Notice that f (−1) = 1. Thus, the slope of the tangent line is given by
f (a + h) − f (a)
(1 − 3h + 3h2 − h3 ) − 1
= lim
= lim −3 = 3h − h2 = −3.
x→−1
x→−1
x→−1
h
h
f 0 (−1) = lim
Hence, the equation to the tangent line is
y = m(x − x1 ) + y1 = −3(x + 1) + 1.
4.
Using the definition of f 0 (a) (by limits), find the slope of the tangent line to f (x) =
and (0, f (0)).
x+1
at (−1, f (−1))
x−1
Solution: First we find f 0 (a):
f 0 (a)
=
=
=
=
=
=
=
lim
x→a
lim
f (x) − f (a)
x−a
x+1
a+1
x−1 − a−1
x−a
(x + 1)(a − 1) − (x − 1)(a + 1)
lim
x→a
(x − 1)(a − 1)(x − a)
2a − 2x
lim
x→a (x − 1)(a − 1)(x − a)
−2(x − a)
lim
x→a (x − 1)(a − 1)(x − a)
−2
lim
x→a (x − 1)(a − 1)
−2
.
(a − 1)2
x→a
Now, for (−1, f (−1)):
1
f 0 (−1) = − .
2
Thus, the equation of the tangent line to f at (−1, f (−1)) is
1
y = f 0 (−1)(x − (−1)) + f (−1) = − (x + 1).
2
For (0, f (0)):
f (0) = −1,
f 0 (0) = −2.
Thus, the equation of the tangent line to f at (0, f (0)) is
f (−1) = 0,
y = f 0 (0)(x − 0) + f (0) = −x − 2.
5.
Find the equation to the tangent line to f (t) = t4 + 2et + 1 at (0, f (0)).
Solution: Using the list of derivatives of basic functions we find that
f 0 (x) = 4t3 + 2ex .
Thus,
f 0 (0) = 2,
f (0) = 3.
Hence, the equation of the tangent line is y = 2x + 3.
6.
Find the points on y = 2x3 + 3x2 − 12x + 1 where the tangent line is horizontal.
Solution: The slope of the tangent line to y = f (x) at any point (a, f (a)) is given by
f 0 (a) = 6x2 + 6x − 12.
We want the tangent line to be horizontal, so we required the slope to be equal to zero. Thus, we need to
solve
0 = f 0 (a) = 6x2 + 6x − 12 = 6(x2 + x − 2) = 6(x + 2)(x − 1).
Thus, a = −2, 1. Hence, at (−2, 21) and (1, −6) the tangent line is horizontal.
7.
Find the point where the tangent line to f (x) = 4ex − 7x + 1 has slope m = −3. Then give the equation of
such tangent line.
Solution: The slope of the tangent line to y = f (x) at any point (a, f (a)) is given by
f 0 (a) = 4ex − 7.
We want the slope of the tangent line to be equal to 3, so we need to solve the equation
−3
=
4ex − 7
4
=
4ex
1
= ex .
Taking natural logarithm on both sides we obtained x = 0.
Since f (0) = 5, we conclude that the equation to the tangent line is y = −3x + 5.
8.
From the top of a cliff a stone is thrown straight up. Assume the stone misses the cliff on the way back
down. The height (with respect to the ground), in meters, at time t, in seconds, is given by
s(t) = −5t2 + 20t + 300.
a)
Find the velocity and acceleration of the stone at time t.
Solution: The velocity is s0 (t) and the acceleration is s00 (t), and we have
s0 (t) = −10t + 20
and
s00 (t) = −10.
b)
Find the height of the cliff. Give the units.
Solution: The height of the cliff is the ”initial hight of the stone”, so the answer is s(0) = 300 meters.
c)
What was its initial velocity? Give the units.
Solution: The initial velocity is s0 (0) = 20 m/sec.
d)
Find its velocity, speed and acceleration at time t = 2. Give the units.
Solution: velocity = s0 (2) = 0 m/sec, so its speed at t = 2 is zero m/sec, while the acceleration is
s00 (2) = −10 m/sec2 .
e)
What was the velocity of the stone when hits the ground? Give the units.
Solution: The stone hits the ground when s(t) = 0. So we need to solve
0 = −5t2 + 20t + 300 = −5(t2 − 4t − 60) = −5(t − 10)(t + 6).
Thus, t = −6, 10. Since time can not be negative, the stone hits the ground after 10 seconds of being
thrown. Thus, the stone hit the ground at a velocity of s0 (10) = −80 m/sec2 .
f)
Find an equation of the tangent line of the graph of s(t) at (4, s(4)).
Solution: Since s(4) = 300 and s0 (4) = −20, the equation of the tangent line is
y = −20(x − 4) + 300.