find the point at which the tangent to the curve

Question

find the point at which the tangent to the curve y=underroot(4x-3) -1 has its slope 2/3.this is the ncert question , after solving this question we get x=3,when we put this value of x in the expression of y we get root(9)-1 . so my doubt is why we do not take +3 and – 3 instead of just +3 as solved in ncert class 12 textbook(applications of derivatives).

Vikas TU · Accepted Answer

Hello Shruti,good qstn..Always remember first that,before doing any such problem which consists any genuine function like,trignometric, modulus, underroot or anything.U must define that function’s domain. .i.e the function or eqn U r dealing with must be in domain of x. Moving to your question here the expression is:y=underroot(4x-3) -1here we have underroot functiion.therfore the function in root must be positive or zero.Hence,4x – 3 > = 0orx > = ¾orx > = 0.75 Hence +3 is > 0.75 rather than -3.

Differential Calculus> find the point at which the tangent to th...

shruti , 8 Years ago

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Differential Calculus> find the point at which the tangent to th...

shruti , 8 Years ago

Vikas TU

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