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

Compute the slope of

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

x = 1 : m = 3

Find the line with slope m=

3

and passing through

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

y = 3 x - 1

\int 7 cos (x) - \frac{4}{1 - x ^{2}} d x

sin (3 t) e^{4 t}

\int sin (\frac{1}{2}) x d x

e^{6}, e^{x}, x = 0

x \to + 2 - lim (\frac{∣ x - x ∣}{x - 2})

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