What is the tangent of f(x)=sqrt(x),(-1,1) ?

The tangent of f(x)=sqrt(x),(-1,1) is y=(sqrt(2)-1)/2 x+1+(sqrt(2)-1)/2

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tangent of f(x)=sqrt(x),(-1,1)

tangent of

f (x) = x, (- 1, 1)

y = \frac{2 - 1}{2} x + 1 + \frac{2 - 1}{2}

Solution steps

Slope of the tangent line at

(- 1, 1) : m = \frac{x - 1}{x + 1}

Find the slope of

f (x) = x : m = \frac{1}{2 x}

Solve

\frac{1}{2 x} = \frac{x - 1}{x + 1} : x = 3 + 22

Find the tangent line at

x = 3 + 22 : y = \frac{2 - 1}{2} x + 1 + \frac{2 - 1}{2}

y = \frac{2 - 1}{2} x + 1 + \frac{2 - 1}{2}

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What is the tangent of f(x)=sqrt(x),(-1,1) ?
The tangent of f(x)=sqrt(x),(-1,1) is y=(sqrt(2)-1)/2 x+1+(sqrt(2)-1)/2