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

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

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

tangent of

3 x^{2} + x, at x = 1

y = 7 x - 3

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

y = 3 x^{2} + x : \frac{d y}{d x} = 6 x + 1

Compute the slope of

y = 3 x^{2} + x

x = 1 : m = 7

Find the line with slope m=

7

and passing through

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

y = 7 x - 3

(- 6, 4), (- 5, - 10)

f (x) = 7 x^{2}, at x = 4

\frac{d}{d x} (\frac{x + 3}{x - 1})

e^{y} \cdot (\frac{d y}{d x} + 1) = 1

y^{^{''}} + 3 y^{^{'}} + 10 y = 0