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How do you find #(f^-1)'(a)# if #f(x)= x/sqrt{x-3}# and a = 4?

Answer 1

Luke Alvarez

The function is not bijective, So, the bi-valued inverse is not differentiable. See the explanation..

For x >= 3, y = f(x) is real.

#y'=1/(x-3)-x/(2(x-3)^(3/2))=(x-6)/(2(x-3)^(3/2))#.

Inversely, bi-valued #x=(y (y+-sqrt(y^2-12)))/2#, with #|y|>=2sqrt3#

Nonetheless, y = 4, x = 4, or 12 are assigned for the two inverse forms.

Both are acceptable.

For the given values of y = a = 4, the inverse is not unique.

distinguishable.

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Answer 2

Emma Aldridge

To find ((f^{-1})'(a)) where (f(x) = \frac{x}{\sqrt{x-3}}) and (a = 4), first, find the inverse of (f(x)) denoted as (f^{-1}(x)). Then differentiate (f^{-1}(x)) with respect to (x) and evaluate the result at (x = 4).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.