Download 1. (a) (2 points) Draw a picture of a function that exactly two local

1. (a) (2 points) Draw a picture of a function that exactly two local minima and one
local maximum.
Solution: I won’t draw a picture, but the formula for a possible answer is f (x) =
x(x − 1)(x − 2)(x − 3). You should draw a picture of this function to convince
yourself it works.
(b) (2 points) Give an example of a function that has a critical point at x = 2, but
does not have a local minimum or a local maximum at that point.
Solution: The simplest example is probably f (x) = (x − 2)3 .
2. (4 points) Use implicit differentiation to write down the tangent line to the curve
y 3 + xy + x3 = 3 at the point (1, 1).
Solution: If we differentiate implicitly with respect to x we get 3y 2 y 0 +y+xy 0 +3x2 = 0.
If we plug in 1 for both x and y we get 3y 0 + 1 + y 0 + 3 = 0 which means that y 0 = −1.
So the tangent line l(x) goes through the point (1, 1) and has slope −1. This means
that l(x) = 1 − 1(x − 1) = 2 − x.
3. (5 points) Given that (tan(x))0 = sec2 (x), use the chain rule to derive a formula for
the derivative of tan−1 (x). (Your ending equation should not have any trigonometric
functions!)
Solution: By definition, we have that
tan(tan−1 (x)) = x.
If we differentiate this equation on the left and right, the chain rule tells us that
tan0 (tan−1 (x))(tan−1 )0 (x) = 1.
But this means that
sec2 (tan−1 (x))(tan−1 )0 (x) = 1.
If we use both the trigonometric identity 1 + tan2 ((tan−1 )0 (x)) = sec2 (tan−1 (x)) and
the fact that tan2 (tan−1 (x)) = x2 , we see that
(1 + x2 )(tan−1 )0 (x) = 1.
This means that (tan−1 )0 (x) =
1
.
1+x2
4. (5 points) Use the definition of the derivative (either “h → 0” or “x → a” is fine) to
show that the derivative of f (x) = 2x3 − x is f 0 (x) = 6x2 − 1.
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Solution: This is basically just a long computation:
f (x + h) − f (x)
h
2(x + h)3 − (x + h) − 2x3 + x
= lim
h→0
h
3
2
2x + 6x h + 6xh2 + 2h3 − x − h − 2x3 + x
= lim
h→0
h
6x2 h + 6xh2 + 2h3 − h
= lim
h→0
h
2
= lim 6x + 6xh + 2h2 − 1
f 0 (x) = lim
h→0
h→0
2
= 6x − 1.
5. (6 points) Find a cubic polynomial f (x) = ax3 + bx2 + cx + d that has a local maximum
value of 3 at x = −2 and a local minimum value of 0 at x = 1. (It might help to draw
a picture.)
Solution: The restrictions from the problem tell us that f (−2) = 3, f 0 (−2) = 0, f (1) =
0, and f 0 (1) = 0. The formula for the derivative is f 0 (x) = 3ax2 + 2bx + c. If we plug
in the 4 restrictions into the functions we get the following equations:
(i) −8a + 4b − 2c + d = 3
(ii) 12a − 4b + c = 0
(iii) a + b + c + d = 0
(iv) 3a + 2b + c = 0
At this point you have to go through and solve the equations by progressively eliminating terms. This is kind of annoying to type out, so I’ll tell you the answer instead:
a = 2/9,
b = 1/3,
c = −4/3,
d = 7/9.
6. (3 points each) Differentiate the following two functions.
Solution:
0
2
0
cos (x)(1 + x2 )
= 2 ln(cos(x)) + ln(1 + x2 ) − ln(1 − x)
ln
1−x
2x
1
= −2 tan(x) +
+
.
2
1+x
1−x
(4 − cos(x))4 (8x + xex )5
0
= 4(4 − cos(x))3 sin(x)(8x + xex )5
+ (4 − cos(x))4 5(8x + xex )4 (8 + ex + xex ).
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