What is the tangent of f(x)=x^3-6x^2-34x+40,\at x=2 ?

The tangent of f(x)=x^3-6x^2-34x+40,\at x=2 is y=-46x+48

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tangent of f(x)=x^3-6x^2-34x+40,\at x=2

tangent of

f (x) = x^{3} - 6 x^{2} - 34 x + 40, at x = 2

y = - 46 x + 48

Solution steps

Find the tangent point:

(2, - 44)

Find the slope of

f (x) = x^{3} - 6 x^{2} - 34 x + 40 : \frac{df ( x )}{d x} = 3 x^{2} - 12 x - 34

Compute the slope of

f (x) = x^{3} - 6 x^{2} - 34 x + 40

x = 2 : m = - 46

Find the line with slope m=

- 46

and passing through

(2, - 44) : y = - 46 x + 48

y = - 46 x + 48

f (x) = x^{2} arcsin (3 x)

f (x) = x^{5} e^{x}

\frac{d r}{d θ} = (sin (π θ))^{4}

\int 3 e^{- \frac{x}{2}} d x

f (x) = 3 e^{x}

What is the tangent of f(x)=x^3-6x^2-34x+40,\at x=2 ?
The tangent of f(x)=x^3-6x^2-34x+40,\at x=2 is y=-46x+48