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2 : Identify points of intersections of graph with parallel lines drawn in the earlier step.
3 : Draw lines of 1 unit parallel to x-axis from intersection points in the direction of positive x. The line ends at the next parallel line on right. Include intersection point but exclude other end of the line. Include transformation for all points of the graph.
The lines drawn in step 3 is the graph of y=f([x]).
Problem : Draw the graph of sin[x].
Solution : Following the construction steps, graph of y=sin[x]is drawn as shown here.
Problem : Draw graph of tan⁻¹[x], x∈[-2, 2].
Solution : Following the construction steps, graph of y= tan⁻¹ [x]is drawn as shown here.
See that function value corresponding to x=2 and x=-2 are not included in the preceding interval on the graph. As such, we need to put a solid circle at x=2 and x=-2 additionally. Further, we need to remove original graph of y= tan⁻¹ x (this step is not shown in the figure above).
The form of transformation is depicted as :
$$y=f\left(x\right)\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}y=\left[f\left(x\right)\right]$$
The graph of y= f(x) is transformed in y=[f(x)] by applying changes to the output of the function. Whatever be the function values, they will be changed to integral values following definition of greatest integer values as given earlier for few intervals. Clearly, real values of “f(x)” are truncated to integer values in the interval of unity i.e. [-1,0), [0,1), [1.2) etc along y-axis.
From the point of construction of the graph of y=f([x]), we need to modify the graph of y=f(x) as :
1 : Draw lines parallel to x-axis (horizontal lines) at integral values along y-axis to cover the graph of y=f(x).
2 : Identify points of intersections of graph with parallel lines drawn in the earlier step. Draw lines parallel to y-axis (vertical lines) from the intersection points identified.
3 : Take x-projection of curve from the point of intersection between two consecutive vertical lines such that it lies on horizontal line of lower value. Include intersection point but exclude other end of the line. Further include points not covered by the projection.
The lines drawn in step 3 is the graph of y=[f(x)].
Problem : Draw the graph of [2sinx].
Solution : Following the construction steps, graph of y=[2sinx]is drawn as shown here.
The form of transformation is depicted as :
$$y=f\left(x\right)\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}\left[y\right]=f\left(x\right)$$
We need to evaluate this equation on the basis of assignment to the dependent expression. The value of function f(x) is first calculated for a given value of x. The value so evaluated is assigned to the GIF function [y]. We interpret assignment to [y]in accordance with the interpretation of equality of the GIF function to a value. In this case, we know that :
$$\left[y\right]=f\left(x\right);\phantom{\rule{1em}{0ex}}f\left(x\right)\notin Z\phantom{\rule{1em}{0ex}}\Rightarrow \text{GIF can not be equated to non-integers. No solution.}$$
$$\left[y\right]=f\left(x\right);\phantom{\rule{1em}{0ex}}f\left(x\right)\in Z\phantom{\rule{1em}{0ex}}\Rightarrow \text{y = Continuous interval of 1 unit starting from f(x)}$$
Clearly, we need to neglect plot corresponding to all non-integral values of f(x). For every value of x, which yields integral value of f(x), there are multiple values of dependent expression [y] in an interval of 1 unit. For example, for $\left[y\right]=f\left(x\right)=\mathrm{2,}y\in 2\le y<3$ . In the nutshell, this graph is not continuous. There is no value of y corresponding to non integer f(x) and there are multiple values of y in an interval of 1 for integral values of f(x).
From the point of construction of the graph of |y|=f(x), we need to modify the graph of y=f(x) as :
1 : Draw lines parallel to x-axis (horizontal lines) at integral values along y-axis to cover the graph of y=f(x).
2 : Identify points of intersections of graph with parallel lines (horizontal lines) drawn in the earlier step.
3 : Draw lines of 1 unit parallel to y-axis (vertical lines) from intersection points in the positive y-direction. Include intersection point but exclude other end of the line.
The lines drawn in step 3 is the graph of [y]= f(x).
Problem : Draw graph of [y]=(x+1)(x-2).
Solution : We first draw the graph of quadratic polynomial function $y=\left(x+1\right)\left(x-2\right)={x}^{2}-x-2$ . The lowest point of the parabola is calculated as :
$$D={\left(-1\right)}^{2}-\left(4X1X-2\right)=1+8=9$$
$${y}_{\mathrm{min}}=-\frac{D}{4a}=-\frac{9}{4X1}=-2.25$$
Following construction steps, graph of [y]=(x+1)(x-2) is drawn as shown here.
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