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

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

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

tangent of

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

y = 4 x - 1

Solution steps

Find the slope of

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

Compute the slope of

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

(1, 3) : m = 4

Find the line with slope m=

4

and passing through

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

y = 4 x - 1

x \to - 1 + lim (f (x))

\int 5 x cos (6 x) d x

x \to 0 lim ((1 + \frac{1}{x})^{3 x})

x \frac{d y}{d x} - 4 y = 6 x^{4} e^{x}

\int_{0}^{5} (22 + t) d t

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