What is the tangent of f(x)=x^2+x,4,\at x=2 ?

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

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tangent of f(x)=x^2+x,4,\at x=2

tangent of

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

y = 5 x - 4

Solution steps

Find the tangent point:

(2, 6)

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

x = 2 : m = 5

Find the line with slope m=

5

and passing through

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

y = 5 x - 4

\int \frac{18}{1 - cos ( 6 x )} d x

sin (t - \frac{( 2 \cdot π )}{3})

(- 3 x + 2)^{4} (- 2 x + 3)^{3}

x \to 0 lim (\frac{2 tan ( \frac{x}{2} )}{x})

\int tan^{5} (6 x) sec^{2} (6 x) d x

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