What is the tangent of f(x)=sqrt(7x+1),\at x=9 ?

The tangent of f(x)=sqrt(7x+1),\at x=9 is y= 7/16 x+65/16

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tangent of f(x)=sqrt(7x+1),\at x=9

tangent of

f (x) = 7 x + 1, at x = 9

y = \frac{7}{16} x + \frac{65}{16}

Solution steps

Find the tangent point:

(9, 8)

Find the slope of

f (x) = 7 x + 1 : \frac{df ( x )}{d x} = \frac{7}{2 7 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{7}{16}

Find the line with slope m=

\frac{7}{16}

and passing through

(9, 8) : y = \frac{7}{16} x + \frac{65}{16}

y = \frac{7}{16} x + \frac{65}{16}

What is the tangent of f(x)=sqrt(7x+1),\at x=9 ?
The tangent of f(x)=sqrt(7x+1),\at x=9 is y= 7/16 x+65/16