Maxima and Minima Analysis: Calculus Assignment Solutions

This assignment solution focuses on the concepts of maxima and minima in calculus, providing detailed analyses of several functions. The solution begins by plotting functions and determining their maxima or minima by differentiating the function and setting it equal to zero. The document then explores the axis of symmetry for parabolas and identifies the zeroes of the functions. It includes the use of the quadratic formula to find zeroes and verifies the results through graphical representations. Furthermore, the solution examines extreme values and restrictions on the variables. Overall, this assignment serves as a comprehensive guide to understanding and solving maxima and minima problems in calculus, offering a clear, step-by-step approach to each problem, including references to relevant mathematical texts.

1Maxima and Minima
Maxima and Minima

2Maxima and Minima
1. Plot of the function : f(x) = 5x2 – x – 5 is given below
-3 -2 -1 0 1 2 3
-10
-8
-6
-4
-2
0
2
4
6
8
10
f(x)
f(x)
According to Hancock(2012), Maxima or minima is calculated by equation the
differentiation of f(x) to 0 i.e.,
df (x )
dx = 0
df (x )
dx = d
dx (5x2 – x – 5) = 0
10x – 1 = 0
x = 0.1
To check maxima or minima, double differentiating f(x)-
d2 f (x)
d x2 |x=0.1 = 10 which is greater than 0. So, x = 0.1 is the point of minima.
Since, the equation given is the equation of parabola, the function is symmetric about
the point of minima or maxima. Here, the minima point is x = 0.1, so the line x = 0.1
is the axis of symmetry.
Also, in the graph, the given line intersects y-axis at x = 0.1 which verifies our result.
2. Plot of the function : f(x) = -2 – x + 2x2 is given below-

3Maxima and Minima
-6 -4 -2 0 2 4 6
-15
-10
-5
0
5
10
15
Column1
Column1
Maxima or minima is calculated by equation the differentiation of f(x) to 0 i.e.,
df (x )
dx = 0
df (x )
dx = d
dx (-2 –x + 2x2) = 0
4x – 1 = 0
x = 0.25
To check maxima or minima, double differentiating f(x)-
d2 f (x)
d x2 |x=0.1 = 4 which is greater than 0. So, x = 0.25 is the point of minima.
Since, the equation given is the equation of parabola, the function is symmetric about
the point of minima or maxima. Here, the minima point is x = 0.25, so the line x =
0.25 is the axis of symmetry.
For zeroes of the function, we need to equate f(x) = 0
-2 – x + 2x2 = 0
As stated in Arfken (2011), for a quadratic equation: ax2 + bx + c = 0, zeroes are given
by : x = −b ± √ b2−4 ac
2a

4Maxima and Minima
x = 1± √(−1)2−4 ( 2 ) (−2)
2(2)
x = 1± √17
4
x = 1.28 or 0.78
In the graph, the given function intersects at x = 1.28 and x = 0.78 which verifies our
zeroes. Also, the function minimises at x = 0.25 which is also the line of symmetry as
shown in the graph.
3. Zeroes : x = -2 and x = +2 are the zeroes of the given function plotted on
graph since the graph cut x-axis at these points.
Extreme values : The maxima of the function is at x = 0 whose value is
y = 0.9(approx.), and the minima of the function is at x = 0.5 whose value is
y = 0.85(approx.)
Restrictions on x and f(x) : x can take all the real values ranging from -∞ to
+∞ except x = -1.5 since f(x) is not defined at x = -1.5. Also, f(x) can also take
any real values ranging from -∞ to +∞ and is not defined only at x = -1.5.

5Maxima and Minima
References-
G. Arfken. Mathematical Methods for Physicists, 7th Edition. Cambridge : Academic
Press, 2011
H. Hancock. Theory of Maxima and Minima. South Carolina : Nabu Press, 2012