Calculus help: using the definition of a derivative. F(x)=√(x-1) and f'(a)=2?

Favorite Answer

Well,

let f(x) = √(x-1)
so
R(x,h) = [ f(x+h) - f(x) ]/h
= [√(x + h - 1) - √(x-1) ]/h
= [√(x + h - 1) - √(x-1) ] * [√(x + h - 1) + √(x-1) ] / ( h * [√(x + h - 1) + √(x-1) ] )
= [ (√(x + h - 1))^2 - (√(x-1))^2 ]/ ( h * [√(x + h - 1) + √(x-1) ] )
= [ x + h - 1 - x + 1]/ ( h * [√(x + h - 1) + √(x-1) ] )
= h/ ( h * [√(x + h - 1) + √(x-1) ] )
= 1 / [√(x + h - 1) + √(x-1) ] --------> 1/ ( 2√(x-1) ) when h ---> 0
therefore for any x =/= -1 :

lim ( h --> 0) [ f(x+h) - f(x) ]/h = f '(x) = 1/ ( 2√(x-1) )

then I will assume you want a so that : f '(a) = 2
this gives us the equation :
1/ ( 2√(a-1) ) = 2
2√(a-1) = 1/2 invert
√(a-1) = 1/4
a - 1 = 1/16

a = 17/16 <--- answer

et voilà, mademoiselle !! ;-)

hope it' ll help !!...Show more

[√[x+h-1] - √(x-1)] / h
([√[x + h - 1] - √(x - 1)] * √[x + h - 1] + √(x - 1)]) / (h[√[x + h -1] + √(x-1)])
[x + h - 1 - (x - 1)] / (h[√[x + h - 1] + √(x - 1)])
h / (h[√[x + h - 1] + √(x - 1)])
1 / [√[x + h - 1] + √(x - 1)]
1 / (√[x + 0 - 1] + √[x - 1])
1 / (√[x - 1] + √[x - 1])
1 / (2√[x - 1])

f'(a)=2
1 / (2√[a - 1]) = 2
1 = 2(2√[a - 1])
1 = 4√[a - 1]
1 / 4 = √[a - 1]
1 / 16 = a - 1
a = 17 / 16...Show more

I made a video on how to go about applying definition of the derivative....Show more

Hint: Rationalize the numerator....Show more