What is the tangent of f(x)=3x^4+2e^x,\at x=0 ?

The tangent of f(x)=3x^4+2e^x,\at x=0 is y=2x+2

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tangent of f(x)=3x^4+2e^x,\at x=0

tangent of

f (x) = 3 x^{4} + 2 e^{x}, at x = 0

y = 2 x + 2

Solution steps

Find the tangent point:

(0, 2)

Find the slope of

f (x) = 3 x^{4} + 2 e^{x} : \frac{df ( x )}{d x} = 12 x^{3} + 2 e^{x}

Compute the slope of

f (x) = 3 x^{4} + 2 e^{x}

x = 0 : m = 2

Find the line with slope m=

2

and passing through

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

y = 2 x + 2

\frac{( s + 3 )}{s ^{2} + s + 3}

\frac{d}{d x} \frac{d y}{d x} - a^{2} y + b = 0

x \to \infty lim (\frac{45}{x})

\frac{d}{d t} (\frac{4}{3} t^{3})

y = x + 1, y = 0, x = 0, x = 2

What is the tangent of f(x)=3x^4+2e^x,\at x=0 ?
The tangent of f(x)=3x^4+2e^x,\at x=0 is y=2x+2