If $y=ƒ(x)$ represents a continuous curve, then the derivative of $ƒ(x)$ with respect to $x$ denoted by $\frac{dy}{dx}$ or $ƒ'(x)$ is the slope of the tangent at any point on the curve. So, the derivative of the given function at $x=x_o$ means the slope of the tangent to the curve at the point $x=x_o$.
To know the nature of the graph of a function, we must have the idea of increasing and decreasing tendency of the function on an interval. Let $y=ƒ(x)$ be the given function which represents a continuous curve. Consider the points $A$, $B$, $C$ and $D$ on the curve where $OP=x_1$, $OQ=x_2$, $OR=x_3$, $OS=x_4$ and $PA=ƒ(x_1)$, $QB=ƒ(x_2)$, $RC=ƒ(x_3)$ and $SD=ƒ(x_4)$.
Here, the curve (i.e. the function) increases from $A$ to $B$ then decreases from $B$ to $C$ and again increases from $C$ to $D$. That is, \[\begin{array}{c} \text{in the part AB, } x_2>x_1 \Rightarrow ƒ(x_2)>ƒ(x_1) \\ \text{in the part BC, } x_3>x_2 \Rightarrow ƒ(x_3)<ƒ(x_2) \\ \text{in the part CD, } x_4>x_3 \Rightarrow ƒ(x_4)>ƒ(x_3) \end{array}\] Hence, we can define the increasing and decreasing functions in the following ways.
Increasing Function
A function $y=ƒ(x)$ is said to be increasing in the interval $(a, b)$ if for every $x_1, x_2\in (a, b)$ \[x_2>x_1\Rightarrow ƒ(x_2)>ƒ(x_1)\] This result shows that as $x$ increases, $y$ i.e. $ƒ(x)$ also increases. So, the slope of the tangent at any point of such a curve is positive. Thus, $y=ƒ(x)$ is an increasing function if $\frac{dy}{dx}=ƒ'(x)>0$.
Decreasing Function
A function $y=ƒ(x)$ is said to be decreasing in the interval $(a, b)$ if for every $x_1, x_2\in (a, b)$ \[x_2>x_1\Rightarrow ƒ(x_2)<ƒ(x_1)\] This result shows that as $x$ increases, $y$ i.e. $ƒ(x)$ decreases. So, the slope of the tangent at any point of such a curve is negative. Thus, $y=ƒ(x)$ is an decreasing function if $\frac{dy}{dx}=ƒ'(x)<0$.
A function is increasing (or decreasing) at the point $x=a$ means the function is increasing (or decreasing) at $x=a$ which is an interior point of an interval.
Examine whether the function $ƒ(x)=15x^2-14x+1$ is increasing or decreasing at $x=\frac{2}{5}$ and $x=\frac{5}{2}$.
\[\text{Here, } ƒ(x)=15x^2-14x+1\] \[ƒ'(x)=30x-14\] \[ƒ’\left(\frac{2}{5}\right)=30×\frac{2}{5}-14=-2<0\] Hence, $ƒ(x)$ is decreasing at $x=\frac{2}{5}$. \[ƒ’\left(\frac{5}{2}\right)=30×\frac{5}{2}-14=61>0\] Hence, $ƒ(x)$ is increasing at $x=\frac{5}{2}$.
Find the intervals in which the function $ƒ(x)=3x^2-6x+5$ is increasing or decreasing.
\[\text{Here, } ƒ(x)=3x^2-6x+5\] \[ƒ'(x)=6x-6\] $ƒ(x)$ will be increasing if \[ƒ'(x)>0\] \[6x-6>0\] \[x-1>0\] \[x>1\] Hence, $ƒ(x)$ is increasing on $(1, \infty)$. And, $ƒ(x)$ will be decreasing if \[ƒ'(x)<0\] \[6x-6<0\] \[x-1<0\] \[x<1\] Hence, $ƒ(x)$ is decreasing on $(-\infty, 1)$.
Find the intervals in which the function $ƒ(x)=6+12x+3x^2-2x^3$ is increasing or decreasing.
\[\text{Here, } ƒ(x)=6+12x+3x^2-2x^3\]\[ƒ'(x)=12+6x-6x^2\] $ƒ(x)$ will be increasing if \[ƒ'(x)>0\] \[12+6x-6x^2>0\] \[6x^2-6x-12<0\] \[x^2-x-2<0\] \[(x+1)(x-2)<0\] \[\text{either, }(x+1)>0 \text{ and } (x-2)<0\] \[x>-1 \text{ and } x<2\] \[\text{or, }(x+1)<0 \text{ and } (x-2)>0\] \[x<-1\text{ and } x>2\]
Hence, $ƒ(x)$ in increasing on $(-1, 2)$. And, $ƒ(x)$ is decreasing on $(-\infty,-1)\cup(2,\infty)$.
