A function \( y = f (x)\) is said to be increasing in an open interval \( (a, b)\) i.e. between the points \( x = a\) and \( x = b\) if \( \frac{dy}{dx} \) or \( f ' (x)\) is positive for all values of \( x\) in that interval. In other words, f(x) is said to be increasing if \( x_1 < x_2 \Rightarrow f (x_1) < f (x_2)\) or equivalently \( x_1 > x_2 \Rightarrow f (x_1) > f (x_2)\) for all real numbers \( x_1\) and \( x_2\) in \( (a, b)\) .
An increasing function can be shown graphically as in the figure below.
The graph of such a function is a curve which goes on rising with the rise in the value of \( x\).
Decreasing Function
A function \( y = f (x)\) is said to be decreasing in an open interval \( (a, b)\) i.e. between the points \( x = a\) and \( x = b\) if \( \frac{dy}{dx} \) or \( f ' (x)\) is negative for all values of \( x\) in that interval. In other words, f(x) is said to be increasing if \( x_1 < x_2 \Rightarrow f (x_1) > f (x_2)\) or equivalently \( x_1 > x_2 \Rightarrow f (x_1) < f (x_2)\) for all real numbers \( x_1\) and \( x_2\) in \( (a, b)\) .
A decreasing function can be shown graphically as in the figure below.
The graph of such a function is a curve which goes on falling with the raise in the value of \( x\).
Example 1
Show that the function \( f(x) = 2x^3 - 3x^2 - 36x \) is increasing on each of the intervals (-∞,-2) and (3,∞), and decreasing on the interval (-2, 3).
Solution Find the derivative. The derivative is \( f'(x) = 6x^2 - 6x - 36\) (1) Factorizing (i) gives \( f'(x) = 6(x^2 - x - 6) = 6(x + 2)(x - 3)\)
(-∞,-2)
(-2,3)
(3,∞)
(x+2)
-
+
+
(x-3)
-
-
+
f'(x)
+
-
+
curve
increasing
decreasing
increasing
When x is less than -2, the values of x + 2 and x - 3 are both negative, and hence the value of \( f'(x) = 6(x + 2)(x - 3) \) is positive. Therefore, by the increasing/decreasing criterion, the function f is increasing on the interval (-∞,-2).
When x is in the interval (-2, 3), the value of x + 2 is positive and the value of x - 3 is negative, and hence the value of \( f'(x) = 6(x + 2)(x - 3)\) is negative. Therefore, by the increasing/decreasing criterion, the function f is decreasing
on the interval (-2, 3).
When x is greater than 3, the values of x + 2 and x - 3 are both positive, and hence the value of \( f'(x) = 6(x + 2)(x - 3)\) is also positive. Therefore, by the increasing/decreasing criterion, the function f is increasing on the interval
(3,∞).
First Derivative Test
Examples : Activity 1
Examine whether the function
\(f(x)=15x^2-14x+1\) is increasing or decreasing at \(x=\frac{2}{5}\) and \(x=\frac{5}{2}\).
Solution
Given function is \(f(x)=15x^2-14x+1\)
Differentiating w.r.to x, we get, \(f'(x)=30x-14\)
The value of \(f'(x)\) at \(x=\frac{2}{5}\) is \(f'(x)=30x-14\)
or\(f'(x)=30 \times \frac{2}{5}-14\)
or\(f'(x)=6 \times 2-14\)
or\(f'(x)=-2\) [NEGATIVE]
It shows that
The function \(f(x)=15x^2-14x+1\) is decreasing at \(x=\frac{2}{5}\)
Next
The value of \(f'(x)\) at \(x=\frac{5}{2}\) is \(f'(x)=30x-14\)
or\(f'(x)=30 \times \frac{5}{2}-14\)
or\(f'(x)=15 \times 5-14\)
or\(f'(x)=61\) [POSITIVE]
It shows that
The function \(f(x)=15x^2-14x+1\) is increasing at \(x=\frac{5}{2}\)
\(f(x)= 2x^3-24x+15 \) is increasing or decreasing at \(x= 3 \) and \(x= \frac{3}{2} \).
Solution
Given function is \(f(x)=2x^3-24x+15\)
Differentiating w.r.to x, we get, \(f'(x)=6x^2-24\)
The value of \(f'(x)\) at \(x=3\) is \(f'(x)=6x^2-24\)
or\(f'(x)=30\) [POSITIVE]
It shows that
The function \(f(x)=2x^3-24x+15\) is increasing at \(x=3\)
Next
The value of \(f'(x)\) at \(x=\frac{3}{2}\) is \(f'(x)=6x^2-24\)
or\(f'(x)=6\times \left (\frac{3}{2} \right )^2-24\)
or\(f'(x)=-\frac{21}{2}\) [NEGATIVE]
It shows that
The function \(f(x)=2x^3-24x+15\) is decreasing at \(x=\frac{3}{2}\)
\(f(x)= 2x^2-4x+3 \) is increasing or decreasing at (1,4).
Solution
Given function is \(f(x)= 2x^2-4x+3\)
Differentiating w.r.to x, we get, \(f'(x)= 4x-4\)
or\(f'(x)= 4(x-1)\)
Now, an analytical table is given below
(1,4)
Factor 1
4
+
Factor 2
(x-1)
+
f'(x)
4(x-1)
+
Result
function is increasing
Therefore, function \(f(x)= 2x^2-4x+3 \) is increasing at (1,4).
\(f(x)= 16x-\frac{4}{3}x^3 \) is increasing or decreasing at \((- \infty ,-2)\).
Solution
Given function is \(f(x)= 16x-\frac{4}{3}x^3\)
Differentiating w.r.to x, we get, \(f'(x)= 16-4x^2\)
or\(f'(x)= 4(4-x^2)\)
or\(f'(x)= 4(2-x)(2+x)\)
Now, an analytical table is given below
(-∞,-2)
Factor 1
4
+
Factor 2
(2-x)
-
Factor 3
(2+x)
+
f'(x)
4(2-x)(2+x)
-
Result
function is decreasing
Therefore, function \(f(x)= 16x-\frac{4}{3}x^3\) is decreasing at (-∞,-2).
Show that the function given by
\( f(x) =- x ^3 + 6x^ 2 -13x+20, x \in R\) is decreasing on R.
Solution
Given function is \(f(x)= - x ^3 + 6x^ 2 -13x+20\)
Differentiating w.r.to x, we get, \(f'(x)= - 3x ^2 + 12x -13\)
or\(f'(x)= - 3x(x - 4) -13\)
Now, an analytical table is given below
(-∞,4)
(4,∞)
Factor 1
-3
-
-
Factor 2
x
-
+
Factor 3
(x - 4)
-
+
f'(x)
- 3x(x - 4)
-
-
Result
function is decreasing
Therefore, function \(f(x)= - x ^3 + 6x^ 2 -13x+20\) is decreasing in R.
\( f(x) = x ^3 – 3x^ 2 + 4x, x \in R\) is increasing on R.
Solution
Given function is \(f(x)= x ^3 – 3x^ 2 + 4x\)
Differentiating w.r.to x, we get, \(f'(x)=3 x ^2 – 6x + 4\)
or\(f'(x)= 3x(x - 2) +4\)
Now, an analytical table is given below
(-∞,2)
(2,∞)
Factor 1
3
+
+
Factor 2
x
-
+
Factor 3
(x - 2)
-
+
f'(x)
3x(x - 2)
+
+
Result
function is decreasing
Therefore, function \(f(x)= x ^3 – 3x^ 2 + 4x\) is increasing in R.
\(f (x) = x^ 3 – 3x^ 2 + 3x – 100\) is increasing in R
Solution
Given function is \(f(x)= 4x-\frac{9}{x}+6\)
Differentiating w.r.to x, we get, \(f'(x)=4+\frac{9}{x^2}\)
The function \(f'(x)=4+\frac{9}{x^2}\) is positive everywhere except x=0
Therefore, \( f(x) = 4x-\frac{9}{x}+6 \) is increasing on R except x=0.
Find the intervals in which the following functions are strictly increasing or decreasing:
Solution
Given function is \(f(x)= x^ 2 – 4x + 6\)
Differentiating w.r.to x, we get, \(f'(x)=2x – 4\)
or \(f'(x)=2(x – 2)\)
Now, an analytical table is given below
(-∞,2)
(2,∞)
Factor 1
2
+
+
Factor 2
(x – 2)
-
+
f'(x)
2(x – 2)
-
+
Result
function is decreasing
function is increasing
The functions is (a) decreasing in (-∞,2) and (b) in increasing in (2,∞)
Solution
Given function is \(f(x)= 4x ^3 – 6x^ 2 – 72x + 30\)
Differentiating w.r.to x, we get, \(f'(x)=12x ^2–12x–72\)
or \(f'(x)=12(x^2-x-6)\)
or \(f'(x)=12(x–3)(x+2)\)
Now, an analytical table is given below
(-∞,-2)
(-2,3)
(3,∞)
Factor 1
(x-3)
-
-
+
Factor 2
(x+2)
-
+
+
f'(x)
12(x–3)(x+2)
+
-
+
Result
increasing
decreasing
increasing
The functions is (a) increasing in (-∞,2) and (b) decreasing in (-2,3) (c) increasing in (3,∞)
\(f(x) =5x ^3 – 135x + 22 \)
\(f(x) =6+12x+3x^2-2x^3 \)
\(f(x) = 2x ^3 – 3x ^2 – 36x + 7\)
\(f (x) = \sin 3x, x \in [0,\pi/2]\)
\(f(x) = \sin x + \cos x, 0 \le x \le 2 \pi \)
Prove that the function given by \(f(x) = \cos x\) is (a) decreasing in (0, π) (b) increasing in (π, 2π), and (c) neither increasing nor decreasing in (0, 2π).
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