What is the tangent of f(2.1)=3x^2-4x+1,(2) ?

The tangent of f(2.1)=3x^2-4x+1,(2) is y=(6a_{0}-4)x-3a_{0}^2+1

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tangent of f(2.1)=3x^2-4x+1,(2)

tangent of

f (2.1) = 3 x^{2} - 4 x + 1, (2)

y = (6 a_{0} - 4) 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} - 4 a_{0} + 1)

Find the slope of

f (2.1) = 3 x^{2} - 4 x + 1 : \frac{df ( 2.1 )}{d x} = 6 x - 4

Compute the slope of

f (2.1) = 3 x^{2} - 4 x + 1

x = a_{0} : m = 6 a_{0} - 4

Find the line with slope m=

6 a_{0} - 4

and passing through

(a_{0}, 3 a_{0}^{2} - 4 a_{0} + 1) : y = (6 a_{0} - 4) x - 3 a_{0}^{2} + 1

y = (6 a_{0} - 4) x - 3 a_{0}^{2} + 1

\int ln ∣ x ∣ cos (x) d x

\frac{\partial}{\partial x} (5 \cdot (y + 4) \cdot (x - 2)^{3})

\frac{\partial}{\partial y} ((x - y)^{2})

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

x \to 2 lim (- x^{3} + 3)

What is the tangent of f(2.1)=3x^2-4x+1,(2) ?
The tangent of f(2.1)=3x^2-4x+1,(2) is y=(6a_{0}-4)x-3a_{0}^2+1