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

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

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

tangent of

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

y = x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

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

Compute the slope of

f (x) = e^{x}

x = 0 : m = 1

Find the line with slope m=

1

and passing through

(0, 1) : y = x + 1

y = x + 1

f (x) = 7 x^{3} + 7 x^{2} + 7 x + 3

\frac{d}{d x} ((x^{4} - 1)^{3} (x^{3} + 1)^{4})

\int_{1}^{2} \frac{1}{8 - 3 t} d t

f (x) = x^{2} - 1, at x = 0

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

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