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

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

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

tangent of

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

Compute the slope of

f (x) = 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

x \to 1 lim (x^{2} + x + 1)

\int_{0}^{4} (x) d x

x \to 0 lim (4 x cos (x))

y^{^{''}} + 81 y = cos (6 t)

f (x) = 5 x - 1, at x = 2

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