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

The tangent of f(x)=3x^3-x^2+2,(1,4) is y=7x-3

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

tangent of

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

y = 7 x - 3

Solution steps

Find the slope of

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

Compute the slope of

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

(1, 4) : m = 7

Find the line with slope m=

7

and passing through

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

y = 7 x - 3

\int t^{3} \cdot ln (t^{2}) d t

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

\frac{\partial}{\partial y} (x e^{- x^{2} y})

\int \frac{5 x ^{4}}{x ^{5} - 3} d x

n = 1 \sum \infty \frac{1}{2 n}

What is the tangent of f(x)=3x^3-x^2+2,(1,4) ?
The tangent of f(x)=3x^3-x^2+2,(1,4) is y=7x-3