What is the tangent of f(x)=2sqrt(7x+1),\at x=9 ?

The tangent of f(x)=2sqrt(7x+1),\at x=9 is y= 7/8 x+65/8

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tangent of f(x)=2sqrt(7x+1),\at x=9

tangent of

f (x) = 2 7 x + 1, at x = 9

y = \frac{7}{8} x + \frac{65}{8}

Solution steps

Find the tangent point:

(9, 16)

Find the slope of

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

Compute the slope of

f (x) = 2 7 x + 1

x = 9 : m = \frac{7}{8}

Find the line with slope m=

\frac{7}{8}

and passing through

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

y = \frac{7}{8} x + \frac{65}{8}

1.5 e^{2 x}

f (x) = x \cdot sin (x)

g (t) = 6 t

\frac{d}{d x} (\frac{2}{x ^{3}} + \frac{5}{x ^{2}})

\int x 4 x - 1 d x

What is the tangent of f(x)=2sqrt(7x+1),\at x=9 ?
The tangent of f(x)=2sqrt(7x+1),\at x=9 is y= 7/8 x+65/8