What is the tangent of f(x)=3x^3-x^2+4,(1,6) ?

The tangent of f(x)=3x^3-x^2+4,(1,6) is y=7x-1

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tangent of f(x)=3x^3-x^2+4,(1,6)

tangent of

f (x) = 3 x^{3} - x^{2} + 4, (1, 6)

y = 7 x - 1

Solution steps

Find the slope of

f (x) = 3 x^{3} - x^{2} + 4 : \frac{df ( x )}{d x} = 9 x^{2} - 2 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 7

Find the line with slope m=

7

and passing through

(1, 6) : y = 7 x - 1

y = 7 x - 1

\frac{d}{d x} (- \frac{6}{5} e^{- 20 x})

\int e^{3 x} - sin (x) + \frac{7}{1 + x ^{2}} d x

\frac{\partial}{\partial x} (ln (x^{2} + 2 y^{2} + 2))

y^{^{'}} t^{4} = 0

\int_{0}^{π} x^{2} sin (x) d x

What is the tangent of f(x)=3x^3-x^2+4,(1,6) ?
The tangent of f(x)=3x^3-x^2+4,(1,6) is y=7x-1