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

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

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

tangent of

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

y = x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

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

Compute the slope of

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

x = 0 : m = 1

Find the line with slope m=

1

and passing through

(0, 0) : y = x

y = x

(1 + x^{2}) (e^{2 t} d t - e^{x} d x) - (1 + x) = 0

x \to π lim (\frac{x}{4} - \frac{π}{2})

x \to 2 lim (\frac{( x - 3 )}{x - 2})

\int_{- 3}^{3} \frac{x}{9 - x ^{2}} d x

x \to 0 lim (\frac{e ^{3 x} - 1}{x})

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