In the previous section, we saw definition 1 and definition 2, which help us to find the nature of a function at any given point. In this section, we will see how derivatives can be applied to find the nature.
To apply derivatives, we need to learn theorem 1. We will first see the theorem. Then we will write an explanation and after that, we will write the proof.
Theorem 1
Let f be continuous in [a,b] and differentiable in (a,b). Then
(a) f is strictly increasing in [a,b] if f'(x) > 0 for each x ∈ (a,b)
(b) f is increasing in [a,b] if f'(x) ≥ 0 for each x ∈ (a,b)
(c) f is strictly decreasing in [a,b] if f'(x) < 0 for each x ∈ (a,b)
(d) f is decreasing in [a,b] if f'(x) ≤ 0 for each x ∈ (a,b)
(e) f is constant in [a,b] if f'(x) = 0 for each x ∈ (a,b)
Explanation can be written in 2 steps:
1. We want to know the nature of the function in the interval [a,b]
2. There are infinite number of points in the interval (a,b).
(a) If the derivative f'(x) is greater than zero at all those points, then we say that:
f is strictly increasing in [a,b].
(b) If the derivative f'(x) is greater than or equal to zero at all those points, then we say that:
f is increasing in [a,b].
(c) If the derivative f'(x) is lesser than zero at all those points, then we say that:
f is strictly decreasing in [a,b]
(d) If the derivative f'(x) is lesser than or equal to zero at all those points, then we say that:
f is decreasing in [a,b]
(e) If the derivative f'(x) is zero at all those points, then we say that:
f is constant in [a,b]
Proof for (a) can be written in 4 steps:
1. Let x1 and x2 be any two points in the interval [a,b], such that x1 < x2.
2. Given that:
♦ f is a real valued function
♦ f is continuous in [a,b]
♦ f is differentiable in (a,b)
•
Then the three conditions are satisfied. We can apply Mean Value Theorem between the two points x1 and x2.
3. There exists a point c between x1 and x2 in such a way that:
Derivative at c is equal to the slope of the secant between x1 and x2.
•
So we get:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{f'(c)} & {~=~} &{\frac{f(x_2) \,-\, f(x_1)}{x_2 \,-\, x_1}} \\
{~\color{magenta} 2 } &{\implies} &{f(x_2) \,-\, f(x_1)} & {~=~} &{f'(c) (x_2 \,-\, x_1)} \\
{~\color{magenta} 3 } &{\implies} &{f(x_2) \,-\, f(x_1)} & {~>~} &{0} \\
{~\color{magenta} 4 } &{\implies} &{f(x_2)} & {~>~} &{f(x_1)} \\
\end{array}$
◼ Remarks:
2 (magenta color):
•
In part (a) of the theorem, it is given that, f'(x) > 0 for all points within (a,b). So f'(c) will be greater than zero.
•
Also, we choose x1 and x2 in such a way that x2 > x1. So (x2 − x1) will be greater than zero.
•
Therefore, f'(c) (x2 − x1) will be greater than zero.
4. From the above step, we see that:
If x1 < x2, then f(x1) < f(x2) and so, the given function is a strictly increasing function.
Proof for (b) can be written in 4 steps:
1. Let x1 and x2 be any two points in the interval [a,b], such that x1 < x2.
2. Given that:
♦ f is a real valued function
♦ f is continuous in [a,b]
♦ f is differentiable in (a,b)
•
Then the three conditions are satisfied. We can apply Mean Value Theorem between the two points x1 and x2.
3. There exists a point c between x1 and x2 in such a way that:
Derivative at c is equal to the slope of the secant between x1 and x2.
•
So we get:
$\begin{array}{ll} {~\color{magenta} 1 }
&{{}} &{f'(c)} & {~=~} &{\frac{f(x_2) \,-\,
f(x_1)}{x_2 \,-\, x_1}} \\
{~\color{magenta} 2 }
&{\implies} &{f(x_2) \,-\, f(x_1)} & {~=~}
&{f'(c) (x_2 \,-\, x_1)} \\
{~\color{magenta} 3 } &{\implies} &{f(x_2) \,-\, f(x_1)} & {~\ge~} &{0} \\
{~\color{magenta} 4 } &{\implies} &{f(x_2)} & {~\ge~} &{f(x_1)} \\
\end{array}$
◼ Remarks:
2 (magenta color):
•
In part (b) of the theorem, it is given that, f'(x) ≥ 0 for all points within (a,b). So f'(c) will be greater than or equal to zero.
•
Also, we choose x1 and x2 in such a way that x2 > x1. So (x2 − x1) will be greater than zero.
•
Therefore, f'(c) (x2 − x1) will be greater than or equal to zero.
4. From the above step, we see that:
If x1 < x2, then f(x1) ≤ f(x2) and so, the given function is an increasing function.
Proof for (c) can be written in 4 steps:
1. Let x1 and x2 be any two points in the interval [a,b], such that x1 < x2.
2. Given that:
♦ f is a real valued function
♦ f is continuous in [a,b]
♦ f is differentiable in (a,b)
•
Then the three conditions are satisfied. We can apply Mean Value Theorem between the two points x1 and x2.
3. There exists a point c between x1 and x2 in such a way that:
Derivative at c is equal to the slope of the secant between x1 and x2.
•
So we get:
$\begin{array}{ll} {~\color{magenta} 1 }
&{{}} &{f'(c)} & {~=~} &{\frac{f(x_2) \,-\,
f(x_1)}{x_2 \,-\, x_1}} \\
{~\color{magenta} 2 }
&{\implies} &{f(x_2) \,-\, f(x_1)} & {~=~}
&{f'(c) (x_2 \,-\, x_1)} \\
{~\color{magenta} 3 } &{\implies} &{f(x_2) \,-\, f(x_1)} & {~<~} &{0} \\
{~\color{magenta} 4 } &{\implies} &{f(x_2)} & {~<~} &{f(x_1)} \\
\end{array}$
◼ Remarks:
2 (magenta color):
•
In part (c) of the theorem, it is given that, f'(x) < 0 for all points within (a,b). So f'(c) will be less than zero.
•
Also, we choose x1 and x2 in such a way that x2 > x1. So (x2 − x1) will be greater than zero.
•
Therefore, f'(c) (x2 − x1) will be less than zero.
4. From the above step, we see that:
If x1 < x2, then f(x1) > f(x2) and so, the given function is a strictly decreasing function.
• Proofs for parts (d) and (e) can be written in a similar way.
Now we will see some solved examples:
Solved example 22.8
Show that the function given by
f(x) = x3 − 3x2 + 4x, x ∈ R is strictly increasing on R.
Solution:
1. f'(x) = 3x2 −6x + 4
2. This can be written as:
f'(x) = 3(x2 −2x + 1) + 1
= 3(x −1)2 + 1
3. There is a square term and a +ve term. So f'(x) will be always greater than zero.
•
We can try input x from any interval (a,b) in R. The value of f'(x) will be never less than zero.
•
Also, since there is a "+ 1" term, f'(x) can never become equal to zero.
Therefore, f is a strictly increasing function.
4. Fig.22.45 below shows the graph of f(x) in red color and f'(x) in yellow color.
 |
| Fig.22.4 |
From the yellow curve we see that, no f'(x) value is zero or less than zero.
In the next section, we will see a few more solved examples.
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