What is the tangent of f(x)=sqrt(x),\at x=4 ?

The tangent of f(x)=sqrt(x),\at x=4 is y= 1/4 x+1

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tangent of f(x)=sqrt(x),\at x=4

tangent of

f (x) = x, at x = 4

y = \frac{1}{4} x + 1

Solution steps

Find the tangent point:

(4, 2)

Find the slope of

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

Compute the slope of

f (x) = x

x = 4 : m = \frac{1}{4}

Find the line with slope m=

\frac{1}{4}

and passing through

(4, 2) : y = \frac{1}{4} x + 1

y = \frac{1}{4} x + 1

ln (2 x - 1) - ln (x - 1)

f (x) = sin (2 x)

y = 3 x + 5

y = \frac{x ^{2}}{x - 1}, (0, 0)

f (x) = (x^{2} - 3 x) ln (x^{2} - 3 x)

What is the tangent of f(x)=sqrt(x),\at x=4 ?
The tangent of f(x)=sqrt(x),\at x=4 is y= 1/4 x+1