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

The tangent of f(x)=sqrt(x+2),\at x=7 is y= 1/6 x+11/6

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

tangent of

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

y = \frac{1}{6} x + \frac{11}{6}

Solution steps

Find the tangent point:

(7, 3)

Find the slope of

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

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}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

(7, 3) : y = \frac{1}{6} x + \frac{11}{6}

y = \frac{1}{6} x + \frac{11}{6}

\int_{0}^{0.2} 600 x d x

- y^{^{''}} + 49 y = 0

\int sin^{2} (t) d t

\frac{3}{s ^{2} - 49}

\frac{d ^{2}}{d x ^{2}} (x 4 - x^{2})

What is the tangent of f(x)=sqrt(x+2),\at x=7 ?
The tangent of f(x)=sqrt(x+2),\at x=7 is y= 1/6 x+11/6