What is the tangent of f(x)=3x^3-6x^2+7x,\at x=4 ?

The tangent of f(x)=3x^3-6x^2+7x,\at x=4 is y=103x-288

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tangent of f(x)=3x^3-6x^2+7x,\at x=4

tangent of

f (x) = 3 x^{3} - 6 x^{2} + 7 x, at x = 4

y = 103 x - 288

Solution steps

Find the tangent point:

(4, 124)

Find the slope of

f (x) = 3 x^{3} - 6 x^{2} + 7 x : \frac{df ( x )}{d x} = 9 x^{2} - 12 x + 7

Compute the slope of

f (x) = 3 x^{3} - 6 x^{2} + 7 x

x = 4 : m = 103

Find the line with slope m=

103

and passing through

(4, 124) : y = 103 x - 288

y = 103 x - 288

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\frac{d}{d x} (\frac{x ^{2} - 11 x + 10}{x ^{2} + 11 x + 10})

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\frac{\partial}{\partial x} (a^{2} - x^{2})

\int \frac{arcsin ( x )}{1 - x} d x

What is the tangent of f(x)=3x^3-6x^2+7x,\at x=4 ?
The tangent of f(x)=3x^3-6x^2+7x,\at x=4 is y=103x-288