Math 20A
Midterm 2 Solutions
1. (a)
f ′ (x) = 6x2 − 6x − 36
(b)
f ′′ (x) = 12x − 6
(c) We get critical points for y = f (x) by setting the first derivative equal to 0.
6x2 − 6x − 36 = 0
x2 − x − 6 = 0
(x − 3)(x + 2) = 0
The critical points are x = 3 and x = −2.
(d) x = 3
f ′′ (3) = 12(3) − 6 = 30 > 0
Since y = f (x) has a horizontal tangent line and is concave up, it must have
a local minimum at x = 3.
x = −2
f ′′ (−2) = 12(−2) − 6 = −30 < 0
Since y = f (x) has a horizontal tangent line and is concave down, it must
have a local maximum at x = −2.
2. (a) We’ll use the identity
y = eln(y)
for y = xln(x) .
xln(x) = eln(x
ln(x)
)
= eln(x) ln(x)
2
= e((ln(x)) )
(b)
d xln(x)
2
= d e((ln(x)) )
2
= e((ln(x)) ) d (ln(x))2
2
= e((ln(x)) ) (2 ln(x)) d(ln(x))
1
2
= 2e((ln(x)) ) ln(x) dx
x
3. (a) We’ll take time derivatives of both sides of the equation, being sure to carefully
use the chain rule.
d
d
(2x2 + y 2 ) =
(18)
dt
dt
dy
dx
+ 2y
= 0
4x
dt
dt
(b)
4(−1)
dx
+ 2(4)(3) = 0
dt
dx
−4
= −24
dt
dx
= 6
dt
(c) Of course, if the train isn’t moving then the x-velocity is equal to the y-velocity
since both are zero. That said, we are interested in the points where the velocities
must be equal due to the geometry of the track, not a stationary train. Therefore,
we’ll assume:
dx
dy
=
6= 0
dt
dt
We then calculate
dx
dy
+ 2y
dt
dt
dx
dx
+ 2y
4x
dt
dt
dx
(4x + 2y)
dt
4x + 2y
2y
y
4x
= 0
= 0
= 0
= 0
= −4x
= −2x
The train must live on the track:
2x2 + y 2
2x2 + (−2x)2
2x2 + 4x2
6x2
x2
And so x =
√
=
=
=
=
=
18
18
18
18
3
√
3 or x = − 3 . Since y = −2x , we have our two points:
√
√
√
√
( 3, −2 3) and (− 3, 2 3)
4. (a) Critical points for y = f (x) are x-values where
x = 1,
x = 3,
(b) The function y = f (x) is increasing where
dy
dx
= f ′ (x) = 0.
x=5
dy
dx
= f ′ (x) is positive.
(1, 3) and (3, 5)
In fact, it’s also correct to say that y = f (x) is increasing on
(1, 5)
or even
[1, 5]
For these solutions, though, we’ll stick to answering using open intervals.
(c) The function y = f (x) is decreasing where
dy
dx
= f ′ (x) is negative.
(0, 1) and (5, 6)
(d) The function y = f (x) is concave up where
dy
must find where dx
= f ′ (x) is increasing.
dy
dx
= f ′′ (x) is positive. That is, we
(0, 2) and (3, 4)
(e) The function y = f (x) is concave down where
dy
dx
= f ′ (x) is decreasing.
(2, 3) and (4, 6)
(f) By the second derivative test, the function y = f (x) has a local max at any
critical point where the function is concave down. Of our three critical points,
only x = 5 satisfies this condition.
We may also find the local max by using the first derivative test. The function
dy
= f ′ (x) has a sign change at x = 5 from positive to negative. This means
dx
that y = f (x) goes from being increasing to decreasing at x = 5.
(g) The function y = f (x) has a local min at x = 1. As with the part (f), we may
understand this using the 2nd derivative test or using the 1st derivative test.
(h) The function y = f (x) is continuous (since it is smooth) and it is defined on a
closed interval [0, 6]. It must therefore have a global max.
We are not asked to find where the global max is located. There are two good
candidates: x = 0 and x = 5. By the end of this course, we’ll have the tools we
need to decide which is the global max.
5. (a)
f (x) = cosh(x)
f ′ (x) = sinh(x)
f ′′ (x) = cosh(x)
f (0) = 1
f ′ (0) = 0
f ′′ (0) = 1
Putting these into the formula for the best quadratic approximation, we get
q(x) = f (0) + f ′ (0) x +
= 1 + 0x +
= 1+
1 2
x
2
1
(1) x2
2
1 ′′
f (0) x2
2
(b)
2
1
1 1
1
1
cosh
≈ q
= 1+
= 1+
2
2
2 2
8
6. We know
Therefore,
dy
= ln(x)
dx
d −1
dx
1
f (y) =
=
dy
dy
ln(x)
Often, we wish to rewrite the right hand side in terms of y by substituting x = f −1 (y).
In this problem, however, we are given an x-value, so we won’t make this substitution.
At x = e2 , we get
1
1
=
2
ln(e )
2