A point on the curve $$y=(x$$−$$3)2$$, where the tangent is parallel to the chord joining the points (3$$,$$0) and (4$$,$$1)

Question

A point on the curve $$y=(x$$−$$3)2$$, where the tangent is parallel to the chord joining the points (3$$,$$0) and (4$$,$$1)

Infinity Learn · Accepted Answer

we have, $$y=(x$$−$$3)2$$, which is polynomial function. So it is continuous and differentiable.  Thus conditions of mean value theorem are satisfied.  Hence, there exists at least one c∈$$3$$,$$4$$  such that,f′(c)=f(4)−$$f(3)4$$−$$3$$⇒$$2(c$$−$$3)=1$$−$$01$$⇒c−$$3=12$$⇒$$c=72$$∈(3$$,$$4)⇒$$x=72$$, where tangent is parallel to the chord joining points (3$$,$$0) and (4$$,$$1) For $$x=72$$,$$y=72$$−$$32=122=14$$ So, $$72$$,$$14$$ is the point on the curve, where tangent drawn is parallel to the chord joining the  points (3$$,$$0) and (4$$,$$1)

$A point on the curve y = (x - 3)^{2}, where the tangent is parallel to the chord joining the points (3, 0) and (4, 1)$

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A point on the curve y=(x−3)2, where the tangent is parallel to the chord joining the points (3,0) and (4,1)

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$A point on the curve y = (x - 3)^{2}, where the tangent is parallel to the chord joining the points (3, 0) and (4, 1)$