What is the tangent of f(x)=sqrt(9-x),\at x=2 ?

The tangent of f(x)=sqrt(9-x),\at x=2 is y=-1/(2sqrt(7))x+sqrt(7)+1/(sqrt(7))

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tangent of f(x)=sqrt(9-x),\at x=2

tangent of

f (x) = 9 - x, at x = 2

y = - \frac{1}{2 7} x + 7 + \frac{1}{7}

Solution steps

Find the tangent point:

(2, 7)

Find the slope of

f (x) = 9 - x : \frac{df ( x )}{d x} = - \frac{1}{2 9 - x}

Compute the slope of

f (x) = 9 - x

x = 2 : m = - \frac{1}{2 7}

Find the line with slope m=

- \frac{1}{2 7}

and passing through

(2, 7) : y = - \frac{1}{2 7} x + 7 + \frac{1}{7}

y = - \frac{1}{2 7} x + 7 + \frac{1}{7}

y^{^{'}} = e^{8 x + y}

\frac{\partial}{\partial y} (z e^{x y})

x \to 1 lim (3 x^{3} - 4 x^{2} + x + 2)

\frac{\partial}{\partial x} (ln (x^{3} + y) - z)

\int \frac{x ^{10}}{( x ^{2} + x - 2 )} d x

What is the tangent of f(x)=sqrt(9-x),\at x=2 ?
The tangent of f(x)=sqrt(9-x),\at x=2 is y=-1/(2sqrt(7))x+sqrt(7)+1/(sqrt(7))