What is the tangent of f(x)=3-6x^2,-57,\at x=3 ?

The tangent of f(x)=3-6x^2,-57,\at x=3 is y=-36x+57

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tangent of f(x)=3-6x^2,-57,\at x=3

tangent of

f (x) = 3 - 6 x^{2}, - 57, at x = 3

y = - 36 x + 57

Solution steps

Find the tangent point:

(3, - 51)

Find the slope of

f (x) = 3 - 6 x^{2} : \frac{df ( x )}{d x} = - 12 x

Compute the slope of

f (x) = 3 - 6 x^{2}

x = 3 : m = - 36

Find the line with slope m=

- 36

and passing through

(3, - 51) : y = - 36 x + 57

y = - 36 x + 57

x \to 3 - lim (ln (x - 3))

\frac{\partial}{\partial x} (x + x y)

\frac{d}{d x} (\frac{2 x ^{\frac{3}{2}} - x ^{2} + 4}{x})

\frac{d}{d x} (\frac{1}{x ^{7}} + \frac{2}{x ^{3}})

\int_{- 1}^{1} 2 - 2 x^{2} d x

What is the tangent of f(x)=3-6x^2,-57,\at x=3 ?
The tangent of f(x)=3-6x^2,-57,\at x=3 is y=-36x+57