Monotonicity, Maxima and Minima
3.0 Monotonicity of a function
3.0 Monotonicity of a function
Monotonicity tells us about the function, whether it is increasing or decreasing in a particular interval.
- A monotonically increasing function is simply an increasing function. Increasing does not mean that the function can not be a constant function. It means that it should not be a decreasing function. As shown in figure, the function is increasing from $A$ to $B$ and then becomes constant from $B$ to $C$ and again increasing from $C$ to $D$. This is a monotonically increasing function.
Let us assume $f$ be any increasing function with points $x_1$ and $x_2$ as shown in figure. The corresponding $y$-coordinates are $f({x_1})$ and $f({x_2})$. A monotonically increasing function must staisfy the condition: If $${x_1} < {x_2}$$ then $$f({x_1}) \leqslant f({x_2})$$ Also from the figure, it is clear that the slope of tangent at any point on the curve is positive i.e., for a monotonically increasing function $$f'(x) \geqslant 0$$
- A monotonically decreasing function is simply a decreasing function. Decreasing does not mean that the function
can not be a constant function. It means that it should not be an increasing function. As shown in figure, the function is decreasing from $A$ to $B$ and then becomes constant from $B$ to $C$ and again decreasing from $C$ to $D$. This is a monotonically decreasing function.
Let us assume $f$ be any decreasing function with points $x_1$ and $x_2$ as shown in figure. The corresponding $y$-coordinates are $f({x_1})$ and $f({x_2})$. A monotonically decreasing function must staisfy the condition: If $${x_1} < {x_2}$$ then $$f({x_1}) \geqslant f({x_2})$$ Also from the figure, it is clear that the slope of tangent at any point on the curve is negative i.e., for a monotonically decreasing function $$f'(x) \leqslant 0$$
- A strictly increasing function is simply an increasing function but the difference is the equals to sign. Let us assume
$f$ be any increasing function with points $x_1$ and $x_2$ as shown in figure. The corresponding $y$-coordinates are $f({x_1})$ and $f({x_2})$. A strictly increasing function must satisfy the condition: If $${x_1} < {x_2}$$ then $$f({x_1}) < f({x_2})$$ Also from the figure, it is clear that the slope of tangent at any point on the curve is positive i.e., for a strictly increasing function $$f'(x) > 0$$
Note:
For a strictly increasing function, further two possibilities are there, depends on the shape of the curve i.e., concave and convex shape.
Let us assume a strictly increasing function $f(x)$ in the interval $x \in (a,b)$, then the curve $y=f(x)$ is concave in $(a,b)$ if tangent drawn at any point on it lies below the curve and the condition for it is $$f''(x) > 0$$
The curve $y=f(x)$ is convex in $(a,b)$ if tangent drawn at any point on it lies above the curve and the condition for it is $$f''(x) < 0$$
Question 5. For any function $y=f(x)$, $f'(x)>0$ and $f''(x)>0$. Two points $x_1$ and $x_2$ are given such that ${x_2} > {x_1}$. Which of the following conditions be true?
(a) $f\left( {\frac{{{x_1} + {x_2}}}{2}} \right) > \frac{{f({x_1}) + f({x_2})}}{2}$
(b) $f\left( {\frac{{{x_1} + {x_2}}}{2}} \right) < \frac{{f({x_1}) + f({x_2})}}{2}$
(c) $f\left( {\frac{{{x_1} + {x_2}}}{2}} \right) = \frac{{f({x_1}) + f({x_2})}}{2}$
Solution: As it is given that $f'(x)>0$, we can say that $f(x)$ is an increasing function. It is also given that $f''(x)>0$, from the
concept explained above, we can say that the graph is of concave shape. Now, let us draw the curve as shown in figure. Joint the two points $A$ and $B$ on the curve with a dotted line as shown. The coordinates of mid-point of $A$ and $B$ is written as $$\frac{{f({x_1}) + f({x_2})}}{2}$$ Finding the mid-point of $x_1$ and $x_2$ and the corresponding coordinates of point $C$ on the curve is written as $$f\left( {\frac{{{x_1} + {x_2}}}{2}} \right)$$ From the marked points $C$ and $D$, we can say that $$f\left( {\frac{{{x_1} + {x_2}}}{2}} \right) < \frac{{f({x_1}) + f({x_2})}}{2}$$ The correct answer is option $(b)$.
- A strictly decreasing function is simply an increasing function but the difference is the equals to sign. Let us assume
$f$ be any decreasing function with points $x_1$ and $x_2$ as shown in figure. The corresponding $y$-coordinates are $f({x_1})$ and $f({x_2})$. A strictly decreasing function must satisfy the condition: If $${x_1} < {x_2}$$ then $$f({x_1}) > f({x_2})$$ Also from the figure, it is clear that the slope of tangent at any point on the curve is negative i.e., for a strictly decreasing function $$f'(x) < 0$$
