What is the tangent of f(x)=sqrt(x+16),\at x=1 ?

The tangent of f(x)=sqrt(x+16),\at x=1 is y= 1/(2sqrt(17))x+sqrt(17)-1/(2sqrt(17))

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tangent of f(x)=sqrt(x+16),\at x=1

tangent of

f (x) = x + 16, at x = 1

y = \frac{1}{2 17} x + 17 - \frac{1}{2 17}

Solution steps

Find the tangent point:

(1, 17)

Find the slope of

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

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}{2 17}

Find the line with slope m=

\frac{1}{2 17}

and passing through

(1, 17) : y = \frac{1}{2 17} x + 17 - \frac{1}{2 17}

y = \frac{1}{2 17} x + 17 - \frac{1}{2 17}

What is the tangent of f(x)=sqrt(x+16),\at x=1 ?
The tangent of f(x)=sqrt(x+16),\at x=1 is y= 1/(2sqrt(17))x+sqrt(17)-1/(2sqrt(17))