What is the tangent of 4x^3-3x^2+4,\at x=2 ?

The tangent of 4x^3-3x^2+4,\at x=2 is y=36x-48

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tangent of 4x^3-3x^2+4,\at x=2

tangent of

4 x^{3} - 3 x^{2} + 4, at x = 2

y = 36 x - 48

Solution steps

Find the tangent point:

(2, 24)

Find the slope of

y = 4 x^{3} - 3 x^{2} + 4 : \frac{d y}{d x} = 12 x^{2} - 6 x

Compute the slope of

y = 4 x^{3} - 3 x^{2} + 4

x = 2 : m = 36

Find the line with slope m=

36

and passing through

(2, 24) : y = 36 x - 48

y = 36 x - 48

\frac{d y}{d x} = 2 y - x^{2}

\int \frac{x ^{2}}{( x - 1 ) ^{3}} d x

\frac{\partial}{\partial x} (\frac{5}{e ^{x}})

y^{^{'}} + 4 y = 3 sin (2 t)

\frac{\partial}{\partial x} (3 x^{4} y^{2} + 2 x y^{2} - 5)

What is the tangent of 4x^3-3x^2+4,\at x=2 ?
The tangent of 4x^3-3x^2+4,\at x=2 is y=36x-48