What is the tangent of f(x)=2e^x+x,\at x=0 ?

The tangent of f(x)=2e^x+x,\at x=0 is y=3x+2

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tangent of f(x)=2e^x+x,\at x=0

tangent of

f (x) = 2 e^{x} + x, at x = 0

y = 3 x + 2

Solution steps

Find the tangent point:

(0, 2)

Find the slope of

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

(0, 2) : y = 3 x + 2

y = 3 x + 2

f (x) = t sin (t)

y = 8 x, y = x^{2}

f (x) = 6 x^{3} - 84 x^{2} + 11000 x

h (x) = (5 - 6 x)^{5}

5 x \frac{d y}{d x} = 2 x e^{x} - 5 y + 6 x^{2}

What is the tangent of f(x)=2e^x+x,\at x=0 ?
The tangent of f(x)=2e^x+x,\at x=0 is y=3x+2