What is the tangent of f(x)=x^2+x,(2,6) ?

The tangent of f(x)=x^2+x,(2,6) is y=5x-4

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tangent of f(x)=x^2+x,(2,6)

tangent of

f (x) = x^{2} + x, (2, 6)

y = 5 x - 4

Solution steps

Find the slope of

f (x) = x^{2} + x : \frac{df ( x )}{d x} = 2 x + 1

Compute the slope of

f (x) = x^{2} + x

(2, 6) : m = 5

Find the line with slope m=

5

and passing through

(2, 6) : y = 5 x - 4

y = 5 x - 4

\frac{d}{d x} (cos^{3} (x) + sin^{3} (x))

x \to \infty lim (\frac{3}{2})

\int_{0}^{π} 2 π x (sin (x)) d x

\int cos (2 + t) d t

(x^{2} + 1) y^{^{'}} + 3 x (y - 1) = 0, y (0) = 5

What is the tangent of f(x)=x^2+x,(2,6) ?
The tangent of f(x)=x^2+x,(2,6) is y=5x-4