Notes to a video lecture on http://www.unizor.com
Algebra+ 10 – Functions
Problem A
Prove that function
f(x) = x/(1+x²) is
(a) monotonously decreasing on interval (−∞,−1),
(b) monotonously increasing on interval (−1,1) and
(c) monotonously decreasing on interval (1,+∞).
Solution A
Our first observation is that, if we replace x with −x, function f(x)=x/(1+x²) changes the sign while retaining the absolute value
f(−x) = −f(x).
That means, our function is odd, and its graph is centrally symmetrical relative to the origin of coordinates.
It also means that f(0)=0 and the graph on interval (−∞,0) can be obtained from the graph on interval (0,+∞) by turning it by 180° around the origin of coordinates.
When this rotation is done, an interval on the positive side of the X-axis, where the function was increasing, would be transformed by a rotation into a corresponding interval on the negative side, and the function there would still be increasing.
Similarly, an interval on the positive side of the X-axis, where the function was decreasing, would be transformed by a rotation into a corresponding interval on the negative side, where the function would still be decreasing.
The following is an illustration of this characteristic of odd functions.
Therefore, it’s sufficient to investigate the behavior of this function on interval (0,+∞), and the results are sufficient to tell how this function behaves on interval (−∞,0).
If we prove that f(x) is monotonously increasing on interval (0,1) from f(0)=0 to f(1)=½, it implies that this function is monotonously increasing on interval (−1,0) from f(−1)=−½ to f(0)=0
Analogously, if we prove that f(x) is monotonously decreasing on interval (1,+∞) from f(1)=½ to infinitesimal value, it implies that this function is monotonously decreasing on interval (−∞,−1) from infinitesimal value to f(−1)=−½.
Let’s prove that f(x) is increasing on interval (0,1).
Assume, both x1 and x2, which is greater than x1, belong to interval (0,1), and prove that f(x2) is greater than f(x1).
Analysis:
x1/(1+x12) < x2/(1+x22)
Multiply both sides by positive value (1+x12)·(1+x12), which does not affect the inequality.
x1·(1+x22) < x2·(1+x12)
Open the parenthesis, getting
x1 + x1·x22 < x2 + x2·x12
Rearrange,
x1·x2·(x2−x1) < x2 − x1
Since we assumed that both x1 and x2, which is greater than x1, belong to interval (0,1), all components of this inequality are positive, we can cancel the multiplier (x2−x1), arriving to an obvious inequality
x1·x2 < 1
Proof:
All transformations of inequalities are invariant and reversible, therefore, from the obvious inequality
x1·x2 < 1
we can obtain the original inequality using only invariant transformations, which proves its validity.
Let’s prove that f(x) is decreasing on interval (1,+∞).
Assume, both x1 and x2, which is greater than x1, belong to interval (1,+∞), and prove that f(x1) is greater than f(x2).
Analysis:
x1/(1+x12) > x2/(1+x22)
Multiply both sides by positive value (1+x12)·(1+x12), which does not affect the inequality.
x1·(1+x22) > x2·(1+x12)
Open the parenthesis, getting
x1 + x1·x22 > x2 + x2·x12
Rearrange,
x1·x2·(x2−x1) > x2 − x1
Since we assumed that both x1 and x2, which is greater than x1, belong to interval (1,+∞), all components of this inequality are positive, we can cancel the multiplier (x2−x1), arriving to an obvious inequality
x1·x2 > 1
Proof:
All transformations of inequalities are invariant and reversible, therefore, from the obvious inequality
x1·x2 > 1
we can obtain the original inequality using only invariant transformations, which proves its validity.
Problem B
Prove that function
f(x) = x + (1/x2) is
(a) monotonously increasing on interval (−∞,0),
(b) monotonously decreasing on interval (0,∛2) and
(c) monotonously increasing on interval (∛2,+∞).
Solution B1
Consider an expression
Δ = f(x2) − f(x1)
for x1 being smaller than x2 on each of the three intervals mentioned in the problem.
Based on the sign of Δ that reflects the relationship between f(x1) and f(x2), we can determine the behavior of the function on each interval.
(a) −∞ < x1 < x2 < 0
Δ = x2+(1/x22) − x1−(1/x12) =
= (x2−x1) + (1/x22−1/x12) =
= (x2−x1) +
+ (x1−x2)·(x1+x2)/(x12·x22) =
Since −∞ < x1 < x2 < 0, all components of the above expression are positive, which proves that in this interval the function is monotonously increasing.
(b) 0 < x1 < x2 < ∛2
Δ = (x2−x1) +
+ (x1−x2)·(x1+x2)/(x12·x22) =
= (x2−x1) ·
· [1 − (x1+x2)/(x12·x22)] =
= (x2−x1) ·
· (x1·x2−1/x1−1/x2)/(x1·x2)
The term (x2−x1) is positive.
The term (x1·x2) is positive.
Let’s evaluate the sign of
x1·x2−1/x1−1/x2 =
= x1·x2 − (1/x1+1/x2)
Component (x1·x2) on interval from 0 to ∛2 does not exceed ∛2·∛2 = ∛4.
Components (1/x1+1/x2) exceeds 1/∛2+1/∛2 = ∛4.
Therefore, in this case Δ is negative, which means that the function is decreasing in this interval.
(c) ∛2 < x1 < x2 < +∞
Evaluating again the term in question
x1·x2 − (1/x1+1/x2)
Now situation with the sign of this term is opposite to the previous one.
Component (x1·x2) on this interval exceed ∛2·∛2 = ∛4.
Components (1/x1+1/x2) does not exceed 1/∛2+1/∛2 = ∛4.
Therefore, in this case Δ is positive, which means that the function is increasing in this interval.
Solution B2
Obviously, using the apparatus of Calculus, the solution is shorter.
The first derivative of the function f(x)=x+(1/x2) is
f'(x) = 1 −2/x3
This derivative is equal to zero only at one point
x = ∛2 = 21/3
The local minimum at that point is
y = 21/3 + 2−2/3
When x is either negative or is greater then ∛2, the derivative is positive and, therefore, the function is increasing.
When x is positive but less than ∛2, the derivative is negative and, therefore, the function is decreasing.
Graph B
Incidentally, the graph of the function f(x)=x+(1/x2) looks like this
