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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).

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).

Grade:12

1 Answers

Vikas TU
14149 Points
4 years ago
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) -1
here we have underroot functiion.
therfore the function in root must be positive or zero.
Hence,
4x – 3 > = 0
or
x > = ¾
or
x > = 0.75
 
Hence +3 is > 0.75 rather than -3.
 

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