What is the tangent of f(x)=4x^2,\at x=0 ?

The tangent of f(x)=4x^2,\at x=0 is y=0

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tangent of f(x)=4x^2,\at x=0

tangent of

f (x) = 4 x^{2}, at x = 0

y = 0

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = 4 x^{2} : \frac{df ( x )}{d x} = 8 x

Compute the slope of

f (x) = 4 x^{2}

x = 0 : m = 0

Find the line with slope m=

0

and passing through

(0, 0) : y = 0

y = 0

7 sin (4 t + 8)

\int \frac{1}{81 e ^{- 5 x} + e ^{5 x}} d x

\int_{0}^{5} y + 4 d y

y^{^{'}} - (\frac{7}{8}) y = - \frac{7}{8} e^{- 7 x} y^{9}

2 c

What is the tangent of f(x)=4x^2,\at x=0 ?
The tangent of f(x)=4x^2,\at x=0 is y=0