<< Chapter < Page | Chapter >> Page > |
From the point of construction of the graph of y=|f(x)|, we need to modify the graph of y=f(x) as :
(i) take the mirror image of lower half of the graph in x-axis
(ii) remove lower half of the graph
This completes the construction for y=|f(x)|.
Problem : Draw graph of $y=\left|\mathrm{cos}x\right|$ .
Solution : We first draw the graph of $y=\mathrm{cos}x$ . Then, we take the mirror image of lower half of the graph in x-axis and remove lower half of the graph to complete the construction of graph of $y=\left|\mathrm{cos}x\right|$
Problem : Draw graph of $y=|{x}^{2}-2x-3|$
Solution : We first draw graph $y={x}^{2}-2x-3$ . The roots of corresponding quadratic equation are -1 and 3. After plotting graph of quadratic function, we take the mirror image of lower half of the graph in x-axis and remove lower half of the graph to complete the construction of graph of $y=|{x}^{2}-2x-3|$
Problem : Draw graph of $y=\left|{\mathrm{log}}_{10}x\right|$ .
Solution : We first draw graph $y={\mathrm{log}}_{10}x$ . Then, we take the mirror image of lower half of the graph in x-axis and remove lower half of the graph to complete the construction of graph of $y=\left|{\mathrm{log}}_{10}x\right|$ .
The form of transformation is depicted as :
$$y=f\left(x\right)\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}\left|y\right|=f\left(x\right)$$
As discussed in the beginning of module, value of function is first calculated for a given value of x. The value so evaluated is assigned to the modulus function |y|. We interpret assignment to |y| in accordance with the interpretation of equality of the modulus function to a value. In this case, we know that :
$$\left|y\right|=f\left(x\right);f\left(x\right)>0\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}y=\pm f\left(x\right)\phantom{\rule{1em}{0ex}}$$
$$\left|y\right|=f\left(x\right);f\left(x\right)=0\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}y=0$$
$$\left|y\right|=f\left(x\right);f\left(x\right)<0\phantom{\rule{1em}{0ex}}\Rightarrow \text{Modulus can not be equated to negative value. No solution}$$
Clearly, we need to neglect all negative values of f(x). For every positive value of f(x), there are two values of dependent expressions -f(x) and f(x). It means that we need to take image of upper part of the graph across x-axis. This is image in x-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 : remove lower half of the graph
2 : take the mirror image of upper half of the graph in x-axis
This completes the construction for |y|=f(x).
Problem : Draw graph of $\left|y\right|=\left(x-1\right)\left(x-3\right)$ .
Solution : We first draw the graph of quadratic function given by $y=\left(x-1\right)\left(x-3\right)$ . Then, we remove lower half of the graph and take mirror image of upper half of the graph in x-axis to complete the construction of graph of $\left|y\right|=\left(x-1\right)\left(x-3\right)$ .
Problem : Draw graph of $$\left|y\right|={\mathrm{tan}}^{-1}x$$ .
Solution : We first draw the graph of function given by $y={\mathrm{tan}}^{-1}x$ . Then, we remove lower half of the graph and take mirror image of upper half of the graph in x-axis to complete the construction of graph of $y={\mathrm{tan}}^{-1}x$ .
The form of transformation is depicted as :
Notification Switch
Would you like to follow the 'Functions' conversation and receive update notifications?