What is the tangent of f(x)=e^x+1,\at x=0 ?

The tangent of f(x)=e^x+1,\at x=0 is y=x+2

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tangent of f(x)=e^x+1,\at x=0

tangent of

f (x) = e^{x} + 1, at x = 0

y = x + 2

Solution steps

Find the tangent point:

(0, 2)

Find the slope of

f (x) = e^{x} + 1 : \frac{df ( x )}{d x} = e^{x}

Compute the slope of

f (x) = e^{x} + 1

x = 0 : m = 1

Find the line with slope m=

1

and passing through

(0, 2) : y = x + 2

y = x + 2

f (x) = \frac{4}{x ^{5}} - \frac{1}{x ^{2}}, at x = 1

y = x^{3} - 13 x^{2} + 48 x + 8

\int_{0}^{1} (x)^{2} d x

\frac{\partial}{\partial x} (5 y)

\frac{d}{d x} y = (x^{2} + 3 x)^{2}

What is the tangent of f(x)=e^x+1,\at x=0 ?
The tangent of f(x)=e^x+1,\at x=0 is y=x+2