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| 1. |
Find the domain and range of the function f(x) is equal to square -9 upon x-3 |
| Answer» Here {tex}f(x) = \\frac{{{x^2} - 9}}{{x - 3}}{/tex}f (x) assume real values for all real values of x except for x - 3 = 0 i.e .x = 3Thus domain of f (x) = R - {3}Let f (x) = y{tex}\\therefore y = \\frac{{{x^2} - 9}}{{x - 3}} = \\frac{{(x + 3)(x - 3)}}{{(x - 3)}}{/tex}{tex} \\Rightarrow {/tex}\xa0y = x + 3y takes all real values except 6 as domain =R-{3}Thus range of f (x) = R - {6}. | |