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

The tangent of f(x)=x^2+2x,\at x=1 is y=4x-1

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

tangent of

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

y = 4 x - 1

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

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

Compute the slope of

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

x = 1 : m = 4

Find the line with slope m=

4

and passing through

(1, 3) : y = 4 x - 1

y = 4 x - 1

\int \frac{1}{b y ^{2} + b ^{3}} d y

\int \frac{1}{x 9 x ^{2} + 64} d x

\int x (3 + 10 x) d x

\frac{d ^{2} y}{d x ^{2}} = 4 y

\frac{d v}{d t} = v^{2} + 5 v + 1

What is the tangent of f(x)=x^2+2x,\at x=1 ?
The tangent of f(x)=x^2+2x,\at x=1 is y=4x-1