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

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

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

tangent of

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

y = - x + 1

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

f (x) = x^{3} - 2 x^{2} + 1 : \frac{df ( x )}{d x} = 3 x^{2} - 4 x

Compute the slope of

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

x = 1 : m = - 1

Find the line with slope m=

- 1

and passing through

(1, 0) : y = - x + 1

y = - x + 1

\frac{d}{d x} (arcsin (e^{2 x}))

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

\frac{d}{d x} (4 x^{2} \cdot sin (4 x))

\int x (x + 2)^{6} d x

\int_{- x}^{x^{2}} e^{t^{2}} d t

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