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

The tangent of f(x)=3e^x+x,\at x=0 is y=4x+3

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tangent of f(x)=3e^x+x,\at x=0

tangent of

f (x) = 3 e^{x} + x, at x = 0

y = 4 x + 3

Solution steps

Find the tangent point:

(0, 3)

Find the slope of

f (x) = 3 e^{x} + x : \frac{df ( x )}{d x} = 3 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 = 4

Find the line with slope m=

4

and passing through

(0, 3) : y = 4 x + 3

y = 4 x + 3

y = \frac{1}{x} + \frac{1}{x ^{2}} + \frac{1}{x ^{3}}

\int - \frac{x}{1} d x

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{3 x} (1 + x)

y = 4^{x} (2 x^{4})

3 (e^{x} + e^{x} x)

What is the tangent of f(x)=3e^x+x,\at x=0 ?
The tangent of f(x)=3e^x+x,\at x=0 is y=4x+3