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

The tangent of f(x)=2x^3-x^2+3x,\at x=1 is y=7x-3

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

tangent of

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

y = 7 x - 3

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

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

Compute the slope of

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

x = 1 : m = 7

Find the line with slope m=

7

and passing through

(1, 4) : y = 7 x - 3

y = 7 x - 3

\int_{0}^{t} e^{x} \cdot sin (π \cdot x) d x

6 e^{x} cos (x)

x \to 6 lim (x^{2} - 36)

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

x \to 0 lim (\frac{( tan ( 5 x ) )}{x})

What is the tangent of f(x)=2x^3-x^2+3x,\at x=1 ?
The tangent of f(x)=2x^3-x^2+3x,\at x=1 is y=7x-3