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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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Solution :We have, `y = x(x-4),x in [0,4]`
(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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