Calculus Assignment: Derivatives, Optimization and Function Analysis

This assignment provides solutions to a set of calculus problems focused on derivatives, maxima, and minima. It includes finding maximum and minimum points of functions, determining values where derivatives equal zero, using logarithmic differentiation, identifying critical numbers, and analyzing function behavior over specific intervals. The problems cover topics such as finding derivatives of various functions (polynomial, composite, logarithmic, and exponential), identifying intervals where functions are increasing or decreasing, and determining concavity. Specific functions analyzed include polynomials, rational functions, and composite functions involving exponentials and logarithms. The solutions involve applying derivative rules, solving equations to find critical points, and using the second derivative test to classify these points. Several graph related questions are also solved using derivative properties.

Mathematics

[b] Image_9382:
Question 1:
[a] lim
x →1
x2−x
x2−1
=lim
x →1
x ( x−1)
(x−1)( x +1)
=lim
x →1
x
x +1
=(1/(1+1))
= ½
[b] lim
h→ 0
( 1+h )2 −1
h
=lim
h→ 0
1+h2 +2 h−1
h
=lim
h→ 0
h (h+2)
h
=2
[c] lim
x →1
1−x
√ x−x
= lim
x →1
(1−x )( √ x +x )
( √ x −x)( √ x+ x )
=lim
x →1
(1−x)( √x +x)
x (1−x)
=lim
x →1
( √ x + x )
x
= 2
[d] lim
x →2
x2−x−2
x−2

=lim
x →2
( x−2)( x+1)
x−2
=2+1
=3
Question 2:
Given f(x) = 2x+1
Therefore,
lim
x →2
f ( x )
= lim
x →2
(2 x+1)
= 5
Now, from the definition of epsilon delta of limit, we can say that with any
challenge ε > 0 for a given f, a, and L. One must answer with a δ >0 such that
0<|x-a|< δ, |f(x)-L|< ε.
Now, in this specific case,
0<|x-2|< δ implies |(2x+1)-5|< ε
Or, |x-2|< ε/2 [simplifying, factoring and dividing by 2 both side]
It immediately gives us
δ = ε/2
Now, given ε = 1
Hence, δ = ½
Question 3:
Part [a]

To formally define a range of values close to a number, we will write the
following:
0 < | x - c | < delta
This represents, in one line instead of an infinite enumeration of input values, a
whole range of values (infinite number of values actually), with the constraints
that
| x - c | < delta; those values are at some distance from the problem point 'c' that
is less than some value 'delta'
0 < | x - c | : none of these values coincides with the problem point itself 'c'
This is the zooming in on the input values (without enumerating them) part. we
can make the range as small as we like.
Part [b]
This is very similar to the first part:
0 < | f(x) - L | < epsilon
This constrains the output values to be closer to some value 'L' than 'epsilon',
this really represents the 'size' of the gap.
Question 4:
Given,
f(x) = (x^2-3x+4)/(x-2)
if we plot the function in graph and try to find out what happens at x =2, we can
see

This clearly indicates that the function has a vertical asymptote at x = 2 [that is
option c]
[c] Image_9383:
Question 1:
Question 2:
[a] given f(x) = x^3 +3x^2 +6
Hence, f’(x) = 3x^2 + 6x = 3x(x+2)
[b] given f(x) = (x-3)^5(x+4)^3
Then, log(f(x)) = 5log(x-3) + 3log(x+4)
Hence, 1/f(x) * f’(x) = 5/(x-3) + 3/(x+4)
Or, f’(x) = f(x)*(5x+20+3x-9)/(x-3)(x+4)
= (x-3)^4(x+4)^2(8x-11)
[c] given f(x) = log(3x+1)
Therefore, f’(x) = 3/(3x+1)
[d] given f(x) = e^(-5x)
Therefore, f’(x) = -5e^(-5x)
[e] given f(x) = e^(g(x)), where g(x) = -x^2/d^2

Therefore, f’(x) = e^g(x)*g’(x)
Now, g’(x) = -2x/d^2
Hence, f’(x) = e^g(x)* -2x/d^2
[f] given f(x) = 1/[4(2x^2lnx-x^2)]
Or, (2x^2lnx-x^2)f(x) =1/4
Applying derivatives,
(2x^2lnx-x^2)f’(x)+f(x)*(4xlnx+2x-2x)=0
Or, x^2(2lnx-1)*f’(x) = -f(x)*4xlnx
Or, f’(x) = -f(x)*(4lnx/x(2lnx-1))
[g] given f(t) = t^3/(1+t^2)^3
Or, log(f(t) = 3[logt-log(1+t^2)]
Therefore, f’(t)/f(t) = 3[1/t – 2t/(1+t^2)]
Or, f’(t) =[3t^3/(1+t^2)^3]*[1/t -2t/(1+t^2)]
[h] given f(x) = (1+2x)^(3/2)
Therefore, f’(x) = (3/2)*(1+2x)^(1/2)*2
=3(1+2x)^(1/2)
[i] given, f(x) = (1+x)/(1-x)
Or, log(f(x)) = log(1+x)-log(1-x)
Therefore, f’(x)/f(x) = 1/(1+x)+1/(1-x)
Or, f’(x) = [(1+x)/(1-x)]*[(1-x+1+x)/(1+x)(1-x)]
= 2/(1-x)^2
Question 3:

Given, f(x) = (x+1)/(x^2+1)
Therefore, f’(x) =((x^2+1)*1-(x+1)*2x)/(x^2+1)^2
Or, f’(x) =(1-x^2)/(1+x^2)^2
Nor, f’(x) =0 implies (1-x)(1+x) =0
x= 1, -1
The function is increasing at [1,∞)
Question 4:
Given e^(-x^2)
Therefore, the curve will look like
The curve is concave upward at (-2,0)

The maximum value is 1.
The value is increasing rapidly within (-1,0) interval.
Question 6:
[a] given ∑
i=1
5
(i+3)
= (1+3)+(2+3)+(3+3)+(4+3)+(5+3)
= 4 + 5 + 6 + 7+ 8
= 30
[b] Given ∑
i=1
5 0
(i)
= 1+ 2+ 3+ …… + 50
=(50)*(50+1)/2
=25*51
=1275
[c] Given ∑
i=1
50
( a )
=a
Question 7:
[a] given y = (x-3)^5(x-4)^5
Or, y =(x^2-7x+12)^5
Therefore, y’ = 5(x^2-7x+12)^4*(2x-7)
Now, y’=0 implies,
5(x^2-7x+12)^4(2x-7)=0

Or, (x-3)^4(x-4)^4(2x-7)=0
It means, x =3,4,3.5
Now, y’’ = 20(x^2-7x+12)^3(2x-7)^2+10(x^2-7x+12)
Therefore, at x = 3
y’’ = 20(9-21+12)^3(6-7)^2+10(9-21+12) =0 so at x =3, the function neither
has minimum point nor maximum point.
Again, at x = 4,
y’’ = 20(16-28+12)^3(8-7)^2+(16-28+12) = 0 so at x=4, the function neither
has minimum point nor maximum point.
Finally, at x = 3.5,
y’’ = 20(12.25-24.5+12)^3(7-7)^2+10(12.25-24.5+12)<0
Hence, at x = 7 the function has maximum point.
[b] given y =(x-3)^5(x-4)^5
Or, log(y) = 5log(x-3)+5log(x-4)
Therefore, y’(1/y) = 5/(x-3)+5/(x-4)
Or, y’ = (x-3)^5(x-4)^5[(5x-20+5x-15)/(x-3)(x-5)]
Or, y’ =(x-3)^4(x-4)^4(10x-35)
Now, y’=0 implies,
(x-3)^4(x-4)^4(10x-35) =0
Or, x = 3, 4, 3.5
Now, y’’ = 20(x^2-7x+12)^3(2x-7)^2+10(x^2-7x+12)
Therefore, at x = 3

y’’ = 20(9-21+12)^3(6-7)^2+10(9-21+12) =0 so at x =3, the function neither
has minimum point nor maximum point.
Again, at x = 4,
y’’ = 20(16-28+12)^3(8-7)^2+(16-28+12) = 0 so at x=4, the function neither
has minimum point nor maximum point.
Finally, at x = 3.5,
y’’ = 20(12.25-24.5+12)^3(7-7)^2+10(12.25-24.5+12)<0
Hence, at x = 7 the function has maximum point.
Question 8:
Given y = x^2/3
Plotting this on graph, we have
The function is concave downward at (-∞,0] and concave upward at [0, ∞).
[d] Image_9384:
Function graphs are showing in red, derivative graphs are showing in blue color.
[a]

Graph of its derivative
[b]
[c]