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

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

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

tangent of

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

y = 2

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

f (x) = 2 x^{3} - 3 x^{2} + 3 : \frac{df ( x )}{d x} = 6 x^{2} - 6 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 = 0

Find the line with slope m=

0

and passing through

(1, 2) : y = 2

y = 2

\frac{1}{2 x ^{6}}

\frac{d}{d x} (2 - 2 x)

\int_{1}^{8} x^{\frac{1}{3}} d x

\int_{0}^{1} \frac{15}{4 y - 1} d y

\frac{d}{d x} (x^{- l n (x)})

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