Calculus Assignment Solution - Derivatives, Integrals, and Limits

This document presents a comprehensive solution to a Calculus assignment, covering a range of topics. The solution begins with linearization problems, including finding linear approximations. It then delves into derivative applications, such as finding critical points, determining intervals of increasing and decreasing functions, and identifying local and global extrema. The Mean Value Theorem is also applied. The assignment further explores concavity, inflection points, and the analysis of function behavior. Integrals are evaluated, and limits are computed using L'Hôpital's rule. The solutions are presented step-by-step, providing clear explanations and calculations for each problem, making it a valuable resource for students studying Calculus.

Solution 1a): Formula for linearization: Given a function, then the formula for
linearization at is,
Solution 1b): Given
So, linear approximation is
Solution 1c): Since, and . Put we get the
approximated value as
Solution 2a: Given that
a): Critical points: Differentiate the given function with respect to x we get
. To find critical points, substitute we get
This implies that
Hence, the critical points are .
b): Differentiate again we get
Now, let’s check the sign of at
At ,
At ,
At ,
At ,
From above calculation, it is observe that function is global maximum and global
minimum at respectively. Now let’s find global maximum and global
minimum values
The global maximum value is and
The global minimum value is

Solution 3a): Mean Value Theorem: Suppose that a function is continuous on
closed interval and differentiable on open interval . Then there exist a real
number such that
Solution 3b): Given .
Since, is continuous in closed and differentiable in open interval . Then
there exist a real number c such that
Simplify further, we get
Hence, the value of c up to two decimal places is
Solution 4a): Given that
Differentiate above expression with respect to x we get,
To find critical points, solve
Hence, critical points are .
Solution 4b): Since, critical points are so the function may increase or decreasing in
the intervals .
Since, on the interval . Hence, the function is increasing in the
interval
Since, on the interval . Hence, the function is decreasing in the interval
.
Solution 4c): Since, this implies that .
At
This implies that the function will attain local maximum and local maximum value is .

At This implies that the function will attain local minimum and local
minimum value is .
Solution 4d): To find interval of concavity, solve we get . Hence,
function may concave up or down in the interval .
Since, in the interval . Hence, function is concave downward in the
interval .
Since, in the interval . Hence, function is concave upward in the
interval .
Solution 4e): The inflection point is
Solution 5:
Integrate both sides
Since,
So
And
Integrate both sides
Since,
Solution: 6a): Given expression . Differentiate with respect to x

Solution: 6bGiven expression . Differentiate with respect to x
Solution:7a): Given limit
. Using L’Hospital’s rule we get
Solution 7b): Given limit
. Using L’Hospital’s rule we get
Solution 7c): Given limit
Solution:7d): Given limit
Simplify the given expression