What is the tangent of f(x)=x^2+x,\at x=3 ?

The tangent of f(x)=x^2+x,\at x=3 is y=7x-9

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tangent of f(x)=x^2+x,\at x=3

tangent of

f (x) = x^{2} + x, at x = 3

y = 7 x - 9

Solution steps

Find the tangent point:

(3, 12)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 7

Find the line with slope m=

7

and passing through

(3, 12) : y = 7 x - 9

y = 7 x - 9

\frac{d}{d x} (- 7 x - 10)

\int 2^{2} d x

\frac{d}{d y} (e^{- 2 y})

\int \frac{ln ^{2} ( y )}{3 y + 8 y ln ^{3} ( y )} d y

\frac{d}{d x} (- x e^{2 x})

What is the tangent of f(x)=x^2+x,\at x=3 ?
The tangent of f(x)=x^2+x,\at x=3 is y=7x-9