What is the tangent of y=x^2+x+3,\at x=3.99 ?

The tangent of y=x^2+x+3,\at x=3.99 is y=8.98x-12.9201

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tangent of y=x^2+x+3,\at x=3.99

tangent of

y = x^{2} + x + 3, at x = 3.99

y = 8.98 x - 12.9201

Solution steps

Find the tangent point:

(3.99, 22.9101)

Find the slope of

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

Compute the slope of

y = x^{2} + x + 3

x = 3.99 : m = 8.98

Find the line with slope m=

8.98

and passing through

(3.99, 22.9101) : y = 8.98 x - 12.9201

y = 8.98 x - 12.9201

\int x sec (2) (x) d x

x, 0, 2

\int x^{\frac{3}{7}} d x

\frac{0.1}{s + 0.1}

\frac{d}{d x} (x z + x y + y z)

What is the tangent of y=x^2+x+3,\at x=3.99 ?
The tangent of y=x^2+x+3,\at x=3.99 is y=8.98x-12.9201