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

The tangent of f(x)=sqrt(x+2),\at x=2 is y= 1/4 x+3/2

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

tangent of

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

y = \frac{1}{4} x + \frac{3}{2}

Solution steps

Find the tangent point:

(2, 2)

Find the slope of

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

Compute the slope of

f (x) = x + 2

x = 2 : m = \frac{1}{4}

Find the line with slope m=

\frac{1}{4}

and passing through

(2, 2) : y = \frac{1}{4} x + \frac{3}{2}

y = \frac{1}{4} x + \frac{3}{2}

x^{3} + x^{2} + \frac{6}{x}

\int \frac{8 x}{x ^{2} - 3} d x

\int \frac{1}{sin ^{2} ( 2 x ) cos ^{2} ( 2 x )} d x

\int sec^{3} (x) - sec (x) d x

What is the tangent of f(x)=sqrt(x+2),\at x=2 ?
The tangent of f(x)=sqrt(x+2),\at x=2 is y= 1/4 x+3/2