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

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

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

tangent of

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

y = 3 x + 3

Solution steps

Find the tangent point:

(0, 3)

Find the slope of

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

Compute the slope of

f (x) = 3 e^{x}

x = 0 : m = 3

Find the line with slope m=

3

and passing through

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

y = 3 x + 3

y^{^{'}} = - 5 y + 5 t^{2} + 2 t, y (0) = \frac{1}{3}

\int \frac{3}{4 - x ^{2}} d x

\frac{d}{d θ} (θ - sin (θ))

x \to 0 lim (\frac{1}{sin ( x )})

\frac{\partial}{\partial x} (\frac{y}{( x + y ) ^{2}})

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