What is the tangent of f(x)=x^3-2x^2+4x+3,\at x=1 ?

The tangent of f(x)=x^3-2x^2+4x+3,\at x=1 is y=3x+3

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tangent of f(x)=x^3-2x^2+4x+3,\at x=1

tangent of

f (x) = x^{3} - 2 x^{2} + 4 x + 3, at x = 1

y = 3 x + 3

Solution steps

Find the tangent point:

(1, 6)

Find the slope of

f (x) = x^{3} - 2 x^{2} + 4 x + 3 : \frac{df ( x )}{d x} = 3 x^{2} - 4 x + 4

Compute the slope of

f (x) = x^{3} - 2 x^{2} + 4 x + 3

x = 1 : m = 3

Find the line with slope m=

3

and passing through

(1, 6) : y = 3 x + 3

y = 3 x + 3

10 x - 8 y \cdot x^{2} + 1 \frac{d y}{d x} = 0, y (0) = 4

f (x) = \frac{( x - 5 )}{( x + 3 ) ^{4}}

\int_{0}^{1} - 12 e^{x} d x

\frac{d}{d x} (x^{6} e^{- x})

ln (x), 6

What is the tangent of f(x)=x^3-2x^2+4x+3,\at x=1 ?
The tangent of f(x)=x^3-2x^2+4x+3,\at x=1 is y=3x+3