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

Question

3275 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-06T08:19:18

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

y = f(x) is real for x >=3.

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

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

However, assigning y = 4, x = 4 or 12.for the two inverse forms.

Both are admissible....

The inverse is not unique for the given y = a = 4. The inverse is not

differentiable.

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