What is the tangent of f(x)=3x^2-1 ?

The tangent of f(x)=3x^2-1 is y=6a_{0}x-3a_{0}^2-1

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tangent of f(x)=3x^2-1

tangent of

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

y = 6 a_{0} x - 3 a_{0}^{2} - 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 a_{0}^{2} - 1)

Find the slope of

f (x) = 3 x^{2} - 1 : \frac{df ( x )}{d x} = 6 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 = 6 a_{0}

Find the line with slope m=

6 a_{0}

and passing through

(a_{0}, 3 a_{0}^{2} - 1) : y = 6 a_{0} x - 3 a_{0}^{2} - 1

y = 6 a_{0} x - 3 a_{0}^{2} - 1

f (x) = 12 x, (4, 24)

y^{^{''}} + 6 y^{^{'}} + 8 y = e^{2 x}

y^{^{''}} - y^{^{'}} - 6 y = 0, y (0) = - 1, y^{^{'}} (0) = 4

\int e^{2 - 5 x} d x

x \to 2 + lim (\frac{1}{x ^{2} - 5 x + 6})

What is the tangent of f(x)=3x^2-1 ?
The tangent of f(x)=3x^2-1 is y=6a_{0}x-3a_{0}^2-1