What is the tangent of f(x)=x^3-4x^2-9x+11 ?

The tangent of f(x)=x^3-4x^2-9x+11 is y=(3a_{0}^2-8a_{0}-9)x-2a_{0}^3+4a_{0}^2+11

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tangent of f(x)=x^3-4x^2-9x+11

tangent of

f (x) = x^{3} - 4 x^{2} - 9 x + 11

y = (3 a_{0}^{2} - 8 a_{0} - 9) x - 2 a_{0}^{3} + 4 a_{0}^{2} + 11

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} - 4 a_{0}^{2} - 9 a_{0} + 11)

Find the slope of

f (x) = x^{3} - 4 x^{2} - 9 x + 11 : \frac{df ( x )}{d x} = 3 x^{2} - 8 x - 9

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} - 8 a_{0} - 9

Find the line with slope m=

3 a_{0}^{2} - 8 a_{0} - 9

and passing through

(a_{0}, a_{0}^{3} - 4 a_{0}^{2} - 9 a_{0} + 11) : y = (3 a_{0}^{2} - 8 a_{0} - 9) x - 2 a_{0}^{3} + 4 a_{0}^{2} + 11

y = (3 a_{0}^{2} - 8 a_{0} - 9) x - 2 a_{0}^{3} + 4 a_{0}^{2} + 11

\int_{- \infty}^{\infty} x^{2} e^{\frac{- x ^{2}}{2}} d x

\frac{d y}{d x}, x y + 4 e^{y} = 4 e

\int \frac{5 - e ^{x}}{e ^{6 x}} d x

\int x^{2} (x^{3} + 5)^{10} d x

y^{^{''}} + 2 y^{^{'}} + y = 3 - e^{- t}

What is the tangent of f(x)=x^3-4x^2-9x+11 ?
The tangent of f(x)=x^3-4x^2-9x+11 is y=(3a_{0}^2-8a_{0}-9)x-2a_{0}^3+4a_{0}^2+11