Calculus Assignment: Derivative Calculations, Graph Analysis, and More

This calculus assignment solution addresses problems related to derivatives and their applications. Question 1 focuses on the chain rule and derivative calculations using given functions and graphs. Question 2 analyzes the truth or falsity of statements about derivatives, requiring the student to provide proofs or counterexamples. Finally, Question 3 involves interpreting a velocity-time graph to determine the farthest distance of a bicyclist from a lake, requiring the calculation of displacement using the area under the curve. The solution demonstrates a strong understanding of calculus concepts, including differentiation rules, graph analysis, and the interpretation of physical applications of derivatives.

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