What is the tangent of f(x)=x^3+3x^2-45x-19 ?

The tangent of f(x)=x^3+3x^2-45x-19 is y=(3a_{0}^2+6a_{0}-45)x-2a_{0}^3-3a_{0}^2-19

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tangent of f(x)=x^3+3x^2-45x-19

tangent of

f (x) = x^{3} + 3 x^{2} - 45 x - 19

y = (3 a_{0}^{2} + 6 a_{0} - 45) x - 2 a_{0}^{3} - 3 a_{0}^{2} - 19

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} + 3 a_{0}^{2} - 45 a_{0} - 19)

Find the slope of

f (x) = x^{3} + 3 x^{2} - 45 x - 19 : \frac{df ( x )}{d x} = 3 x^{2} + 6 x - 45

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 3 a_{0}^{2} + 6 a_{0} - 45

Find the line with slope m=

3 a_{0}^{2} + 6 a_{0} - 45

and passing through

(a_{0}, a_{0}^{3} + 3 a_{0}^{2} - 45 a_{0} - 19) : y = (3 a_{0}^{2} + 6 a_{0} - 45) x - 2 a_{0}^{3} - 3 a_{0}^{2} - 19

y = (3 a_{0}^{2} + 6 a_{0} - 45) x - 2 a_{0}^{3} - 3 a_{0}^{2} - 19

y = x^{3} - 2 x + 2 (3.23)

3 x^{2}

\frac{d}{d x} (\frac{1}{2 x} + \frac{1}{7 x ^{\frac{6}{7}}})

\frac{\partial}{\partial x} (ln (4 y e^{x y}))

x \to 3 + lim (\frac{x - 3}{x ^{2} - 9})

What is the tangent of f(x)=x^3+3x^2-45x-19 ?
The tangent of f(x)=x^3+3x^2-45x-19 is y=(3a_{0}^2+6a_{0}-45)x-2a_{0}^3-3a_{0}^2-19