What is the tangent of f(x)=1+sqrt(4-x),\at x=-2 ?

The tangent of f(x)=1+sqrt(4-x),\at x=-2 is y=-1/(2sqrt(6))x+1+sqrt(6)-1/(sqrt(6))

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tangent of f(x)=1+sqrt(4-x),\at x=-2

tangent of

f (x) = 1 + 4 - x, at x = - 2

y = - \frac{1}{2 6} x + 1 + 6 - \frac{1}{6}

Solution steps

Find the tangent point:

(- 2, 1 + 6)

Find the slope of

f (x) = 1 + 4 - x : \frac{df ( x )}{d x} = - \frac{1}{2 - x + 4}

Compute the slope of

f (x) = 1 + 4 - x

x = - 2 : m = - \frac{1}{2 6}

Find the line with slope m=

- \frac{1}{2 6}

and passing through

(- 2, 1 + 6) : y = - \frac{1}{2 6} x + 1 + 6 - \frac{1}{6}

y = - \frac{1}{2 6} x + 1 + 6 - \frac{1}{6}

\int \frac{4 x y - 2 x ^{2}}{2 x ^{3} - x ^{2} y} d y

\int 90 (2 x - 5)^{8} d x

n = 5 \sum \infty 4 - \frac{4}{n ^{2}}

y^{^{'}} = \frac{2 t}{3 y ^{2}}

y^{^{'}} = 14 x y

What is the tangent of f(x)=1+sqrt(4-x),\at x=-2 ?
The tangent of f(x)=1+sqrt(4-x),\at x=-2 is y=-1/(2sqrt(6))x+1+sqrt(6)-1/(sqrt(6))