What is the tangent of f(x)=sqrt(x+2),\at x=1 ?

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

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tangent of f(x)=sqrt(x+2),\at x=1

tangent of

f (x) = x + 2, at x = 1

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

Solution steps

Find the tangent point:

(1, 3)

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 = 1 : m = \frac{1}{2 3}

Find the line with slope m=

\frac{1}{2 3}

and passing through

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

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

f (x) = lo g_{4} (3 x^{2} + 1)

x \to 14 - lim (15 x + ln (14 - x))

(1 + e^{- t})^{2}

y = x^{s i n (x)}

t^{2} - 8 t + 16

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