What is the tangent of f(x)=sqrt(4-x),\at x=0 ?

The tangent of f(x)=sqrt(4-x),\at x=0 is y=-1/4 x+2

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tangent of f(x)=sqrt(4-x),\at x=0

tangent of

f (x) = 4 - x, at x = 0

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

Solution steps

Find the tangent point:

(0, 2)

Find the slope of

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

Compute the slope of

f (x) = 4 - x

x = 0 : m = - \frac{1}{4}

Find the line with slope m=

- \frac{1}{4}

and passing through

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

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

\frac{\partial}{\partial x} (x y - \frac{2}{x} - \frac{4}{y} + 8)

\frac{\partial}{\partial v} (u + v)

\frac{x}{x ^{2} 1}

\frac{d}{d x} (- \frac{2}{( 1 + x ) ^{\frac{3}{2}}})

f (x) = \frac{7}{x + 1}, at x = 7

What is the tangent of f(x)=sqrt(4-x),\at x=0 ?
The tangent of f(x)=sqrt(4-x),\at x=0 is y=-1/4 x+2