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

The tangent of f(x)=3x^2,\at x=1 is y=6x-3

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

tangent of

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

y = 6 x - 3

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

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

Compute the slope of

f (x) = 3 x^{2}

x = 1 : m = 6

Find the line with slope m=

6

and passing through

(1, 3) : y = 6 x - 3

y = 6 x - 3

\frac{d}{d x} (x (4 x + 1)^{\frac{1}{3}})

\int \frac{t}{t - 2} d t

\int \frac{1}{x - c} d x

\int \frac{x ^{4} - 2 x ^{2} + 1}{x ^{3}} d x

x \to 1 - lim (4 x - 5)

What is the tangent of f(x)=3x^2,\at x=1 ?
The tangent of f(x)=3x^2,\at x=1 is y=6x-3