What is the tangent of f(x)=sqrt(2x+1),\at x=3 ?

The tangent of f(x)=sqrt(2x+1),\at x=3 is y= 1/(sqrt(7))x+sqrt(7)-3/(sqrt(7))

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tangent of f(x)=sqrt(2x+1),\at x=3

tangent of

f (x) = 2 x + 1, at x = 3

y = \frac{1}{7} x + 7 - \frac{3}{7}

Solution steps

Find the tangent point:

(3, 7)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{1}{7}

Find the line with slope m=

\frac{1}{7}

and passing through

(3, 7) : y = \frac{1}{7} x + 7 - \frac{3}{7}

y = \frac{1}{7} x + 7 - \frac{3}{7}

What is the tangent of f(x)=sqrt(2x+1),\at x=3 ?
The tangent of f(x)=sqrt(2x+1),\at x=3 is y= 1/(sqrt(7))x+sqrt(7)-3/(sqrt(7))