Math 220
April 9
I.Find the derivative of the function.
1.
x
Z
t2 dt
f (x) =
0
2.
x
Z
f (x) =
tan(s)ds
2
3.
3
Z
2 +5
eu
f (x) =
du
x
4.
x2
Z
sin(t3 + 5)dt
f (x) =
4
5.
Z
x4 +x2
es cos(s)ds
f (x) =
sin(x)
II. Show that the equation has exactly one real root.
x3 + e x = 0
III.
(a) Show the a polynomial of degree 3 has at most three real roots.
(b) Show that a polynomial of degree n has at most n real roots.
IV. Sketch the region enclosed by the given curves and find its area.
1.
y = 12 − x2 ,
1
y = x2 − 6
2.
y = x2 ,
y=
√
x
3.
y = x3 ,
y=x
4.
y = sin(x),
y = 2x/π
5.
x = y2,
x=3
6.
x = y2,
7.
y=
√
x − 2,
8.
y=
9.
y=
√
x,
x=y+6
√
x,
x−y =2
y = x/3
y = x/3,
x = 16
10.
y = |x|,
x2 − 6
y = |x|,
6 − x2
11.
12.
y = 1/x,
y = 4x,
y = x/4
13.
x = y 2 − 4,
x = 4 − y2
14.
x = y 2 − y − 6,
2
y = x/6 + 3
1
Solutions
I.Find the derivative of the function.
1.
x
Z
t2 dt
f (x) =
0
Answer:
f 0 (x) = x2
2.
x
Z
tan(s)ds
f (x) =
2
Answer:
f 0 (x) = tan(x)
3.
3
Z
2 +5
eu
f (x) =
du
x
Answer:
f 0 (x) = −ex
4.
2 +5
x2
Z
sin(t3 + 5)dt
f (x) =
4
Answer:
f 0 (x) = 2x sin(x6 + 5)
5.
Z
x4 +x2
es cos(s)ds
f (x) =
sin(x)
Answer:
f 0 (x) = (4x3 + 2x)ex
4 +x2
cos(x4 + x2 ) − cos(x)esin(x) cos(sin(x))
3
II. Show that the equation has exactly one real root.
x3 + e x = 0
Answer:
Let f (x) = x3 + ex , assume f (x) has two roots, that is f (r1 ) = f (r2 ) = 0.
The mean value theorem states since f is continuous and differiable
There exist c ∈ (r1 , r2 ) such that:
f 0 (x) =
f (r2 ) − f (r1 )
0−0
=
=0
r2 − r1
r2 − r1
However, f 0 (x) = 3x2 + ex > 0 for all x. Contradiction. Therefore f (x)
cannot has two roots and can have at most root.
Since f (0) >) and f (−10) < 0, by the intermediate value theorem states
that exist c ∈ (−10, 0) such that f (c) = 0. Thus f (x) has exactly one root.
III.
(a) Show the a polynomial of degree 3 has at most three real roots.
Answer:
Let p(x) be a polynomial of degree 3 with four roots. The mean value
theorem states that p0 (x) will have 3 roots. However p0 (x) will have
degree 2 and therefore will have at most 2 roots. So we get a contradiction, which means p(x) cannot exist. Thus a polynomial of degree
3 has at most three real roots.
(b) Show that a polynomial of degree n has at most n real roots.
Answer:
Let p(x) be a polynomial of degree N , where N is the lowest degree
where a polynomial has more roots than the value of its degree. Then
p0 (x) will have more than N − 1 roots by the mean value theorem. Also
p0 (x) will have degree N − 1, which is a contradiction. So p(x) does not
exist. Therefore a polynomial of degree n has at most n real roots.
value theorem we
IV. Sketch the region enclosed by the given curves and find its area.
4
1.
y = 12 − x2 ,
y = x2 − 6
Answer:
10
5
-4
2
-2
4
-5
Z
3
[(12 − x2 ) − (x2 − 6)]dx
Area =
−3
2.
y = x2 ,
y=
√
x
Answer:
1.2
1.0
0.8
0.6
0.4
0.2
0.2
0.4
0.6
0.8
1
Z
Area =
√
[( x) − (x2 )]dx
0
3.
y = x3 ,
5
1.0
y=x
Answer:
1.0
0.5
-1.0
0.5
-0.5
1.0
-0.5
-1.0
Z
0
Z
3
[x − x]dx +
Area =
−1
1
[x − x3 ]dx
0
4.
y = sin(x),
y = 2x/π
Answer:
1.0
0.5
-1.5
-1.0
0.5
-0.5
1.0
1.5
-0.5
-1.0
Z
0
Z
[2x/π − sin(x)]dx +
Area =
−π/2
[sin(x) − 2x/π]dx
0
5.
x = y2,
6
π/2
x=3
Answer:
1.5
1.0
0.5
0.5
1.0
1.5
2.0
2.5
3.0
-0.5
-1.0
-1.5
2
Z
Area =
√
√
[ x − − x]dx
0
6.
x = y2,
x=y+6
Answer:
2
2
4
6
8
-2
-4
-6
Z
3
[(y + 6) − y 2 ]dx
Area =
−2
7.
y=
√
x − 2,
7
x−y =2
Answer:
1.0
0.8
0.6
0.4
0.2
2.0
2.2
2.4
2.6
2.8
3.0
-0.2
Z
Area =
3
√
[ x − 2 − (x − 2)]dx
2
8.
y=
√
x,
y = x/3
6
8
Answer:
3.0
2.5
2.0
1.5
1.0
0.5
2
4
9
Z
Area =
√
[ x − (x/3)]dx
0
9.
y=
√
x,
y = x/3,
8
x = 16
Answer:
5.0
4.5
4.0
3.5
3.0
10
12
14
Z
16
1
6[(x/3) −
Area =
√
x]dx
9
10.
y = |x|,
x2 − 6
Answer:
2
-3
-2
1
-1
2
3
-2
-4
-6
Z
0
Z
2
[(−x) − (x − 6)]dx +
Area =
−3
0
11.
y = |x|,
9
6 − x2
3
[(x) − (x2 − 6)]dx
Answer:
6
4
2
-3
-2
1
-1
2
3
-2
Z
0
Z
2
[(6 − x ) − (−x)]dx +
Area =
−3
3
[(6 − x2 ) − (x)]dx
0
12.
y = 1/x,
y = 4x,
y = x/4
Answer:
2.0
1.5
1.0
0.5
0.0
0.5
1.0
Z
1.5
1/2
2.0
Z
2
[4x − (x/4)]dx +
Area =
0
[(1/x − (x/4)]dx
1/2
13.
x = y 2 − 4,
10
x = 4 − y2
Answer:
3
2
1
0
-1
-2
-3
-4
0
-2
2
Z
4
2
Area =
[(4 − y 2 ) − (y 2 − 4)]dy
−2
14.
x = y 2 − y − 6,
11
y = x/6 + 3
Answer:
4.0
3.8
3.6
3.4
3.2
3.0
0
1
2
3
Z
Area =
4
5
6
4
[(6y − 18) − (y 2 − y − 6)]dy
3
12