What is the tangent of f(x)=x^2+x,\at (x=1) ?

The tangent of f(x)=x^2+x,\at (x=1) is y=3x-1

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tangent of f(x)=x^2+x,\at (x=1)

tangent of

f (x) = x^{2} + x, at (x = 1)

y = 3 x - 1

Solution steps

Find the tangent point:

(1, 2)

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 = 3

Find the line with slope m=

3

and passing through

(1, 2) : y = 3 x - 1

y = 3 x - 1

What is the tangent of f(x)=x^2+x,\at (x=1) ?
The tangent of f(x)=x^2+x,\at (x=1) is y=3x-1