What is the tangent of f(x)=-6sqrt(x),\at x=4 ?

The tangent of f(x)=-6sqrt(x),\at x=4 is y=-3/2 x-6

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tangent of f(x)=-6sqrt(x),\at x=4

tangent of

f (x) = - 6 x, at x = 4

y = - \frac{3}{2} x - 6

Solution steps

Find the tangent point:

(4, - 12)

Find the slope of

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

Compute the slope of

f (x) = - 6 x

x = 4 : m = - \frac{3}{2}

Find the line with slope m=

- \frac{3}{2}

and passing through

(4, - 12) : y = - \frac{3}{2} x - 6

y = - \frac{3}{2} x - 6

\int 3 sin^{4} (x) d x

x^{4} - 3 x^{2} + 2

\frac{d}{d x} (cos (x^{2} + 2 x))

\frac{8 x ^{2} + 4 x + 4}{x}

x = y^{2} + 6 y, x = y

What is the tangent of f(x)=-6sqrt(x),\at x=4 ?
The tangent of f(x)=-6sqrt(x),\at x=4 is y=-3/2 x-6