The equation of the tangent line at x = 1 is given by:
y = 2x - 1
The equation of the normal line at x = 1 is given by:
y = -1/2x + 3/2
Find the equation of the tangent and normal line of the equation:
f(x) = √x^2 - 1, at x = 1
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Show your complete solution.
3 answers
You must have meant
f(x) = √(x^2 - 1) or else you would have written f(x) = x - 1
f ' (x) = (1/2)(x^2 - 1)^(-1/2) (2x)
= x/√(x^2 - 1)
at x = 1 , f'(1) = 1/0 , which is undefined
so we must have a vertical tangent at (1, 0)
and the equation of that tangent is x = 1
the normal would be a horizontal line, y = 0
f(x) = √(x^2 - 1) or else you would have written f(x) = x - 1
f ' (x) = (1/2)(x^2 - 1)^(-1/2) (2x)
= x/√(x^2 - 1)
at x = 1 , f'(1) = 1/0 , which is undefined
so we must have a vertical tangent at (1, 0)
and the equation of that tangent is x = 1
the normal would be a horizontal line, y = 0
geez @Stefan, will you stop posting this same lame problem?