What is the tangent of f(x)=sqrt(2x-1),\at x=5 ?

The tangent of f(x)=sqrt(2x-1),\at x=5 is y= 1/3 x+4/3

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tangent of f(x)=sqrt(2x-1),\at x=5

tangent of

f (x) = 2 x - 1, at x = 5

y = \frac{1}{3} x + \frac{4}{3}

Solution steps

Find the tangent point:

(5, 3)

Find the slope of

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

Compute the slope of

f (x) = 2 x - 1

x = 5 : m = \frac{1}{3}

Find the line with slope m=

\frac{1}{3}

and passing through

(5, 3) : y = \frac{1}{3} x + \frac{4}{3}

y = \frac{1}{3} x + \frac{4}{3}

\int_{0}^{1} 11 cos (x^{2}) d x

\frac{3}{5 x ^{3}}

\int x^{2} 4 + x^{2} d x

f (x) = 2 x^{2} - 5 x + 4, at x = 0

n = 1 \sum \infty \frac{3 n}{4 n + 5}

What is the tangent of f(x)=sqrt(2x-1),\at x=5 ?
The tangent of f(x)=sqrt(2x-1),\at x=5 is y= 1/3 x+4/3