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

The tangent of f(x)=(2x-1)^3,\at x=1 is y=6x-5

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

tangent of

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

y = 6 x - 5

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

f (x) = (2 x - 1)^{3} : \frac{df ( x )}{d x} = 6 (2 x - 1)^{2}

Compute the slope of

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

x = 1 : m = 6

Find the line with slope m=

6

and passing through

(1, 1) : y = 6 x - 5

y = 6 x - 5

\int x^{- 2} e^{- x} d x

y^{^{''}} + 10 y^{^{'}} + 34 y = 0

x d y - y d x = 2 x^{2} y^{2} d y, y (1) = 2

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

\int \frac{8 - 3 x}{25 - 64 x ^{2}} d x

What is the tangent of f(x)=(2x-1)^3,\at x=1 ?
The tangent of f(x)=(2x-1)^3,\at x=1 is y=6x-5