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

The tangent of f(x)=2x^2,\at x=1 is y=4x-2

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

tangent of

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

y = 4 x - 2

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 4

Find the line with slope m=

4

and passing through

(1, 2) : y = 4 x - 2

y = 4 x - 2

\int 5 x^{3} + 3 x + 7 d x

\frac{d}{d t} (a cos (t))

x \to 0 lim (- \frac{ln ( 1 + 6 x )}{x})

\int (4 x^{6} - 2 x^{3} + 7 x - 4) d x

\frac{\partial}{\partial x} (\frac{x}{r})

What is the tangent of f(x)=2x^2,\at x=1 ?
The tangent of f(x)=2x^2,\at x=1 is y=4x-2