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(2))x+1+sqrt(2)+1/(sqrt(2))

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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 2} x + 1 + 2 + \frac{1}{2}

Solution steps

Find the tangent point:

(2, 1 + 2)

Find the slope of

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

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 2}

Find the line with slope m=

- \frac{1}{2 2}

and passing through

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

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

x \to \infty lim ((x) 3)

cos (2 x), 0

\frac{\partial}{\partial t} (r cos (t) sin (s))

\frac{d}{d x} (x (x - 4)^{3})

(\frac{x}{x ^{3} - 1})^{^{'}}

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(2))x+1+sqrt(2)+1/(sqrt(2))