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

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

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

tangent of

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

y = - 3 x + 2

Solution steps

Find the tangent point:

(1, - 1)

Find the slope of

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

Compute the slope of

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

x = 1 : m = - 3

Find the line with slope m=

- 3

and passing through

(1, - 1) : y = - 3 x + 2

y = - 3 x + 2

\int \frac{1}{x ^{2} + 6 x + 10} d x

\int 33 sec^{4} (x) d x

y^{^{'}} = t^{2} y

y^{^{'}} = y + 2

\int \frac{( ( ln ( x ) ) ^{5} )}{x} d x

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