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Using Rolle's theroem, find the point on the curve y = x (x-4), xin [0,4], where the tangent is parallelto X-axis. |
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Answer» (i) y is a continuous function since`x(x-4)` is a polynomialfunction. Hence, ` y = x (x-4)` iscontinuousin `[0,4]`. (ii) `y'= (x-4).1+x.1=2x-4` which existsin `(0,4)`. Hence, `y` is differentiablein `(0,4)`. (iii) `y(0) = 0(0-4) = 0` and `y(4) = 4(4-4) = 0` ` rArr y(0) =y(4)` Since,conditions of Rolle's theorem are satisfied. Hence, there exists a POINT such that `f(c) = 0` in `(0,4), [:' f(x) = y']` `rArr 2c - 4 = 0` `rArr c = 2` `rArr x =2, y = 2(2-4) = - 4` THUS, `(2,-4)`is the pointon the curve at which thetangent drawnis parallel to X-axis. |
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