Math1220 Midterm 1 Review Problems
Answer Key
1.
2.
3.
y=−4x+4+ π
2
3
1+5 √ x
f −1( x)=
3
2−2 √ x
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
4.
t=
y'=
2cos (3x )(−sin (3x))(3)
3
+
2
cos (3x )
√1−(3x−2)2
5(2x 2 )
)
5x +3
y ' =π (1+ x 4 )π−1 (4x 3 )+π1+x (ln π)(4x 3 )
y ' =−sech(cos (2x)) tanh(cos (2x ))(−sin (2x ))(2)
3
y'=
−12x−7+12x 2−5cos(5x )
3x−2
1
−1
1
y ' =e 3x
+ 3x (−3)
2
3x
e
3x 2 (ln x )
1
y ' =( x 3 −1)ln x ln (x 3 −1)+ 3
x
x −1
−sin x
y'=
√( cos x+3)2−1
2
y ' =(5x +3)2x ( 4x ln (5x+3)+
4
( )
(
−10 ln(0.08)
years
ln 2
5. Evaluate each integral.
1
√ 1−4x 2+C
2
5 ln∣2x 2 +x−7∣+C
−5arctan( ln x)+C
4−4−5
ln 16
1 2
( y arctan ( y)− y +arctan ( y )) +C
2
−1 √ 3/2
( 2 −2 )
ln 2
−1
ln 4
3
6 4−1
ln 36
1
ln (e 2x +5)+C
2
(a) x arcsin(2x )+
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
)
5
(arcsin ( x 3 ))+C
3
1
x2
(k)
arctan
+C
4
2
1
ln ( x 4 +4)+C
(l)
4
3
4x
x
(m)
x (4 )−
ln 4
ln 4
π
(n) − 2
(j)
( )
(
6.
1
14
7.
f ( x )=
)
sin x+1
and since sin x is always between -1 and 1, then 1+sin x must
cos x
be between 0 and 2 (inclusive) which is always nonnegative. The denominator is also always
positive wherever the function actually exists. This means that the derivative is always
nonnegative in the entire domain of the function. This is the same as saying that it's
monotonic.
8.
9.
1
4
(
f ' ( x)=−
2
+15( x−1)2
1+4x 2
)
which is always positive inside the parentheses since
all coefficients are positive and the powers on x are even. Thus, the derivative is always
negative which means the inverse function exists.
10.
( )
3 5x 15
=e
x
x →∞
5x
(b) lim ( 1 ) =1
(a) lim 1+
x →∞
( f −1 )' (11)=
1
−1
=
f ' (0) 17