What is the tangent of f(x)=x^3-3x^2-9x+27,\at x=4 ?

The tangent of f(x)=x^3-3x^2-9x+27,\at x=4 is y=15x-53

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tangent of f(x)=x^3-3x^2-9x+27,\at x=4

tangent of

f (x) = x^{3} - 3 x^{2} - 9 x + 27, at x = 4

y = 15 x - 53

Solution steps

Find the tangent point:

(4, 7)

Find the slope of

f (x) = x^{3} - 3 x^{2} - 9 x + 27 : \frac{df ( x )}{d x} = 3 x^{2} - 6 x - 9

Compute the slope of

f (x) = x^{3} - 3 x^{2} - 9 x + 27

x = 4 : m = 15

Find the line with slope m=

15

and passing through

(4, 7) : y = 15 x - 53

y = 15 x - 53

\frac{\partial}{\partial x} (\frac{3 x + y ^{2}}{x y})

\int (5 x^{4} - 8 x^{3} + 9 x^{2} - 2 x + 7) d x

y^{^{'}} = y^{5}

\frac{d}{d x} (\frac{3 x - 5}{2 x + 3})

x \to 0 lim ((sin (\frac{x}{2}))^{x})

What is the tangent of f(x)=x^3-3x^2-9x+27,\at x=4 ?
The tangent of f(x)=x^3-3x^2-9x+27,\at x=4 is y=15x-53