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

The tangent of f(x)=sqrt(x+7),\at x=9 is y= 1/8 x+23/8

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

tangent of

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

y = \frac{1}{8} x + \frac{23}{8}

Solution steps

Find the tangent point:

(9, 4)

Find the slope of

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

Compute the slope of

f (x) = x + 7

x = 9 : m = \frac{1}{8}

Find the line with slope m=

\frac{1}{8}

and passing through

(9, 4) : y = \frac{1}{8} x + \frac{23}{8}

y = \frac{1}{8} x + \frac{23}{8}

\int tan (θ) 4 sin (2 θ) d θ

\int \frac{7}{( z - 10 ) ^{7}} d z

\int_{- 1}^{1} 3^{x} d x

f (x) = 18 x, (9, 54)

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

What is the tangent of f(x)=sqrt(x+7),\at x=9 ?
The tangent of f(x)=sqrt(x+7),\at x=9 is y= 1/8 x+23/8