What is the tangent of f(x)=sqrt(2x+45),\at x=2 ?

The tangent of f(x)=sqrt(2x+45),\at x=2 is y= 1/7 x+47/7

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tangent of f(x)=sqrt(2x+45),\at x=2

tangent of

f (x) = 2 x + 45, at x = 2

y = \frac{1}{7} x + \frac{47}{7}

Solution steps

Find the tangent point:

(2, 7)

Find the slope of

f (x) = 2 x + 45 : \frac{df ( x )}{d x} = \frac{1}{2 x + 45}

Compute the slope of

f (x) = 2 x + 45

x = 2 : m = \frac{1}{7}

Find the line with slope m=

\frac{1}{7}

and passing through

(2, 7) : y = \frac{1}{7} x + \frac{47}{7}

y = \frac{1}{7} x + \frac{47}{7}

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What is the tangent of f(x)=sqrt(2x+45),\at x=2 ?
The tangent of f(x)=sqrt(2x+45),\at x=2 is y= 1/7 x+47/7