How do you find the tangent line approximation to $f (x) = e^{x}$ $f(x)=e^x$ near $x = 0$ $x=0$ ?

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Answer 1 · 2014-09-24T04:19:59

We have to find the derivative of $f (x)$ $f(x)$ and the input the x-value into the derivative to find the numeric slope.

$f (x) = e^{x}$ $f(x)=e^x$

$f (0) = e^{0} = 1$ $f(0)=e^0=1$

Point of tangency is $(0, 1)$ $(0,1)$ .

Note that we actually use the chain rule to find the derivative of $e^{x}$ $e^x$ .

$f'(x)=e^x*1=e^x$

$f'(0)=e^0=1 => slope or m$

How do you find the tangent line approximation to f(x)=e^xf(x)=exf(x)=e^x near x=0x=0x=0 ?