What is the tangent of f(x)=-3x^2+1x,\at x=0 ?

The tangent of f(x)=-3x^2+1x,\at x=0 is y=x

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tangent of f(x)=-3x^2+1x,\at x=0

tangent of

f (x) = - 3 x^{2} + 1 x, at x = 0

y = x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

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

Compute the slope of

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

x = 0 : m = 1

Find the line with slope m=

1

and passing through

(0, 0) : y = x

y = x

\frac{d}{d x} (1 + 3 sin^{3} (x))

x^{2} + 1 + x^{2}

\int_{\frac{π}{4}}^{x} sec^{2} (t) d t

\int \frac{x + 1}{x ^{2} - 2 x - 1} d x

\int (2 t - 1)^{2} d t

What is the tangent of f(x)=-3x^2+1x,\at x=0 ?
The tangent of f(x)=-3x^2+1x,\at x=0 is y=x