What is the tangent of f(x)=sqrt(6-x),\at x=-3 ?

The tangent of f(x)=sqrt(6-x),\at x=-3 is y=-1/6 x+5/2

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tangent of f(x)=sqrt(6-x),\at x=-3

tangent of

f (x) = 6 - x, at x = - 3

y = - \frac{1}{6} x + \frac{5}{2}

Solution steps

Find the tangent point:

(- 3, 3)

Find the slope of

f (x) = 6 - x : \frac{df ( x )}{d x} = - \frac{1}{2 6 - x}

Compute the slope of

f (x) = 6 - x

x = - 3 : m = - \frac{1}{6}

Find the line with slope m=

- \frac{1}{6}

and passing through

(- 3, 3) : y = - \frac{1}{6} x + \frac{5}{2}

y = - \frac{1}{6} x + \frac{5}{2}

\frac{d}{d x} (x^{- 2} e^{x})

x^{2} - 4 x - 6, - x^{2} + 4 x + 4

\frac{d y}{d x} = y^{2} \cdot x^{- 3}

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

\int 9 x e^{- 4 x} d x

What is the tangent of f(x)=sqrt(6-x),\at x=-3 ?
The tangent of f(x)=sqrt(6-x),\at x=-3 is y=-1/6 x+5/2