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Draw the graph of y=2x^(2)-1 and heance the graph of f(x)=cos^(-1)2x^(2)-1). |
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Answer» Solution :`y=2x^(2)-1` is an UPWARD parabola having VERTEX at (0,-1). It meets the x-axis at `(+-sqrt0)` For `f(x)=cos^(-1)(2x^(2)-1)`to get defined. `-1le2x^(2)-1le1` `or""0le2x^(2)le2` `or""0lex^(2)le1` `or""-1lexle1` Hence the domain of `y=f(x) is [-1,1]` `Now f(-1)=f(1)=cos^(-1)=0" and "f(0)=cos^(-1)(-1)=pi` So the inportant points of the graph paper are as FOLLOWS. `"Now"f'(x)=(4X)/(sqrt(1-(2x^(2)-1))^(2))` Clearly f'(x) does not EXIST at x=0, heance f(x) is non-differenctiable at x=0 `" Also"f'(x)gt0 "for x in [-1,0]` `"and"f'(x)lt0 for x in (0, 1]`. Hence the graph of `y=cos^(-1)(2x^(2)-1)`can be drawn as follows.
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