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

The tangent of f(x)=3x^2+x,\at x=1 is y=7x-3

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

tangent of

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

y = 7 x - 3

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

f (x) = 3 x^{2} + x : \frac{df ( x )}{d x} = 6 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 = 7

Find the line with slope m=

7

and passing through

(1, 4) : y = 7 x - 3

y = 7 x - 3

sin^{2} (t)

\frac{d}{d x} (\frac{x}{2 x + 3})

\int_{1}^{1 - 2 y} \frac{y}{x} d x

\frac{d}{d x} (9 x^{2} + y^{2})

y = 5 x^{2}, (4, 80)

What is the tangent of f(x)=3x^2+x,\at x=1 ?
The tangent of f(x)=3x^2+x,\at x=1 is y=7x-3