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

The tangent of f(x)=x^2+x,\at x=4 is y=9x-16

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

tangent of

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

y = 9 x - 16

Solution steps

Find the tangent point:

(4, 20)

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 = 4 : m = 9

Find the line with slope m=

9

and passing through

(4, 20) : y = 9 x - 16

y = 9 x - 16

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = 1

\frac{d}{d x} (\frac{- 6 x}{x ^{2} + 1})

9 x^{2} y^{^{'}} = y^{^{'}} + 5 x e^{- y}

\frac{d y}{d x} = x \cdot e^{- 2 x}

\int \frac{x - 1}{( x + 1 ) ( 2 x ^{2} + 3 )} d x

What is the tangent of f(x)=x^2+x,\at x=4 ?
The tangent of f(x)=x^2+x,\at x=4 is y=9x-16