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= 1/2 x+3/2

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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} x + \frac{3}{2}

Solution steps

Find the tangent point:

(- 1, 1)

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}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

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

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

x \to 1 + lim (x^{2} - 3)

\frac{d}{d x} (- 6 e^{5 x})

\int (ln (4 x)) d x

\int tan^{8} (x) sec^{6} (x) d x

\frac{d}{d t} (1 + t)

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= 1/2 x+3/2