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

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

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tangent of f(x)=sqrt(x),\at x=1

tangent of

f (x) = x, at x = 1

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

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

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

Compute the slope of

f (x) = x

x = 1 : m = \frac{1}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

(1, 1) : y = \frac{1}{2} x + \frac{1}{2}

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

y = 2^{x}

x = 1

(3, 3)

f (x) = \frac{3}{x}

(- 4, 4)

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