What is the tangent of f(x)=sqrt(9-4x),\at x=-1 ?

The tangent of f(x)=sqrt(9-4x),\at x=-1 is y=-2/(sqrt(13))x+sqrt(13)-2/(sqrt(13))

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tangent of f(x)=sqrt(9-4x),\at x=-1

tangent of

f (x) = 9 - 4 x, at x = - 1

y = - \frac{2}{13} x + 13 - \frac{2}{13}

Solution steps

Find the tangent point:

(- 1, 13)

Find the slope of

f (x) = 9 - 4 x : \frac{df ( x )}{d x} = - \frac{2}{9 - 4 x}

Compute the slope of

f (x) = 9 - 4 x

x = - 1 : m = - \frac{2}{13}

Find the line with slope m=

- \frac{2}{13}

and passing through

(- 1, 13) : y = - \frac{2}{13} x + 13 - \frac{2}{13}

y = - \frac{2}{13} x + 13 - \frac{2}{13}

\frac{d}{d x} (\frac{9 x}{4})

f (x) = 9 - x^{2}, g (x) = - 8 x

\int x^{5} d x

T

\frac{x}{cos ( x )}

What is the tangent of f(x)=sqrt(9-4x),\at x=-1 ?
The tangent of f(x)=sqrt(9-4x),\at x=-1 is y=-2/(sqrt(13))x+sqrt(13)-2/(sqrt(13))