What is the tangent of f(x)=x^3-6x^2-34x+40,\at x=4 ?

The tangent of f(x)=x^3-6x^2-34x+40,\at x=4 is y=-34x+8

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tangent of f(x)=x^3-6x^2-34x+40,\at x=4

tangent of

f (x) = x^{3} - 6 x^{2} - 34 x + 40, at x = 4

y = - 34 x + 8

Solution steps

Find the tangent point:

(4, - 128)

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 = 4 : m = - 34

Find the line with slope m=

- 34

and passing through

(4, - 128) : y = - 34 x + 8

y = - 34 x + 8

\frac{\partial}{\partial x} (9 x sin (6 x^{2} y))

y = \frac{5 x}{1 + x ^{2}}, (5, \frac{25}{26})

f (x) = (\frac{x + 5}{x ^{2} + 2})^{2}

1 0^{x^{2} - 1}

What is the tangent of f(x)=x^3-6x^2-34x+40,\at x=4 ?
The tangent of f(x)=x^3-6x^2-34x+40,\at x=4 is y=-34x+8