What is the tangent of f(x)=sqrt(7-6x),\at x=-1 ?

The tangent of f(x)=sqrt(7-6x),\at x=-1 is y=-3/(sqrt(13))x+sqrt(13)-3/(sqrt(13))

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tangent of f(x)=sqrt(7-6x),\at x=-1

tangent of

f (x) = 7 - 6 x, at x = - 1

y = - \frac{3}{13} x + 13 - \frac{3}{13}

Solution steps

Find the tangent point:

(- 1, 13)

Find the slope of

f (x) = 7 - 6 x : \frac{df ( x )}{d x} = - \frac{3}{7 - 6 x}

Compute the slope of

f (x) = 7 - 6 x

x = - 1 : m = - \frac{3}{13}

Find the line with slope m=

- \frac{3}{13}

and passing through

(- 1, 13) : y = - \frac{3}{13} x + 13 - \frac{3}{13}

y = - \frac{3}{13} x + 13 - \frac{3}{13}

\frac{d}{d x} (ln (tan (\frac{x}{2} + \frac{π}{4})))

\int 9 e^{3 t} d t

\frac{d}{d x} (lo g_{5} (2 x - 5))

f (x) = \frac{5}{x}, at x = \frac{1}{4}

\int (y - x) d x

What is the tangent of f(x)=sqrt(7-6x),\at x=-1 ?
The tangent of f(x)=sqrt(7-6x),\at x=-1 is y=-3/(sqrt(13))x+sqrt(13)-3/(sqrt(13))