How do you use the first and second derivatives to sketch $y = e^(1-2x)$ ?

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Answer 1 · 2017-12-15T06:01:09

See argument below.

$y=e^(1-2x)$

$dy/dx = d/dx e^(1-2x)$

$dy/dx = e^(1-2x) * d/dx (1-2x)$

$dy/dx=-2 e^(1-2x)$

Similarly, $(d^2y)/dx^2 = +4e^(1-2x)$

For extrema points of $y, dy/dx =0$

However, in this case, $-2 e^(1-2x) <0 forall x in RR$

Consider, $lim_(x->+oo) -2 e^(1-2x) =0 and lim_(x->+oo) e^(1-2x) =0$

Also consider, $(d^2y)/dx^2 =+4e^(1-2x) >0 forall x in RR$

Hence, it seems reasonable to deduce that $y$ tends to its minimum value of $0$ as $x$ tends to $+oo$

This helps us visualise the graph of $y$ below.

The other important point we need for the graph is at $x=0$

Here, $y = e^(1-0) = e$

We see the point $(0,e)$ on the graph below.

graph{e^(1-2x) [-5.69, 5.406, -1.722, 3.827]}

How do you use the first and second derivatives to sketch y = e^(1-2x)y = e^(1-2x)?