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

The tangent of f(x)=(3x-1)^2,\at x=1 is y=12x-8

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

tangent of

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

y = 12 x - 8

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

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

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 = 12

Find the line with slope m=

12

and passing through

(1, 4) : y = 12 x - 8

y = 12 x - 8

x \to 2 lim (x^{2} - x - 6)

\frac{\partial}{\partial x} (2 x^{3} - 10 y)

\int_{6}^{8} \frac{y}{y ^{2} - 2 y - 3} d y

\int (x^{3} - 1)^{4} x^{2} d x

n = 1 \sum \infty \frac{1}{n ( n + 2 )}

What is the tangent of f(x)=(3x-1)^2,\at x=1 ?
The tangent of f(x)=(3x-1)^2,\at x=1 is y=12x-8