Solutions to Calculus Problems

This page contains solutions to calculus problems related to derivatives, concavity, and displacement. It includes step-by-step explanations and calculations for each problem. The problems cover topics such as finding derivatives, determining concavity, and calculating displacement.

Question 1
a) j' ( 4 )
j ( x )=f ( g ( x ) )
j ( 4 )=f ( g ( 4 ) )
j' ( 4 ) =g' ( f ' ( 4 ) )
g' ( x ) =1
2
j' ( 4 )= 1
2∗1= 1
2
b) k' ( 2 )
k ( x )=g ( f ( x ) )
k' ( 2 )=g' (f ' ( 2 ) )=1
2∗3= 3
2
c) m' ( 1 )
From the graph above,
h ( x )=−|x −2|+4
h' ( x )= −x−2
¿ x−2∨¿ When x=1 , h' ( x )=1 ¿
m ( x ) =g ( f ( h ( x ) ) )
m' ( 1 )=g ' ( f ' ( h' ( 1 ) ) )
¿ 1
2∗2∗1=1
d) n' ( 3 )
n ( x )= f ( x )
h ( x )

n' ( x ) = f ' ( x )
h' ( x )
when x=3. h' ( x )=−1
n' ( 3 )= 4
−1 =−4
e) p' (1)
p ( x )=f ( x )∗h ( x )
f ' ( 1 ) =2
h' ( 1 )=1
p' ( 1 ) =2∗1=2
Question 2
a) The statement is false.
Let f ( x ) =2 x2−16 x +35
f ' ( x ) = 2 ( x +h )2 −16 ( x+ h ) + 35−2 x2−16 x+ 35
h
¿ 2 x2 +4 xh+2h2−16 x−16 h+35−2 x2 +16 x−35
h
¿ 4 xh+2 h2 −16 h
h =4 x +2 h−16
Thisimplies that f ' ( x )=4 x+ 2h−16 … which is false .
b) The statement is false.
Let f ( x ) =x2 +10
f ' ( x )=2 x
g ( x )=x2+100

g' ( x ) =2 x
Thus although f ( x ) ≠ g ( x ) ,
f ' ( x ) =g' (x )
c) The statement is false.
let f ( x )=x4 f ' ( x )=4 x3
f '' ( x )=12 x2
f '' ( a ) =f ' ' ( 0 )=0
f ( a )=0
Thereafter we need ¿ verify that concavity is different on either sid e of x=0
f '' (−1 )=12∗ (−1 )2=12
f '' ( 1 )=12∗ (1 )2 =12
Since the second derivative is positive on either side of x = 0, then the concavity is up on both
sides and x = 0 is not an inflection point. Hence the statement is false.
Question 3
The bicyclist is farthest from the lake when she has been going backwards for a while and speed
decelerate o zero but has turned around started going forward. To get the distance between her
and the lake, we calculate the estimated area below the curve from t=0 when she is farthest from
the lake. The net displacement from the lake will be given by the sum of the negative and
positive values. Thereafter, adding absolute value of the sum to 10 miles will give the
approximate displacement farthest from the lake.
Each square measures 5 miles by 1
6 hr . Giving an approximate area of 31
6 .

On the positive side , there are approximately 4 full squares hence giving 31
6 ∗4=62
3
Onthe negative side ,there are approximately 10 full squares hence giving 31
6 ∗10=−155
3
The Net displacement =31 miles.
Thus the approximate farthest distance=31+10=41 miles