Mathematics, 27.08.2019 10:00 babygreg2001p97abr

Let f(x)= (1+e^x)^-2
i need:
x and y intercepts
derivative and critical #s
local min and max
intervals on concavity and inflection points
any asymptotes
. you so much

Answers: 2

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Answer from: mydoggy152

X-intercepts occur where f(x)=0

, and y-intercept occur where x=0

.

$(1+e^x)^{-2}=\dfrac1{(1+e^x)^2}=0$

has no solution because the denominator is always positive for all real values of

. Thus there are no x-intercepts.

When x=0

, you have $f(0)=\dfrac1{(1+e^0)^2}=\dfrac14$ , so f(x)

intercepts the y-axis at $\left(0,\dfrac14\right)$ .

The derivative is

$\dfrac{\mathrm d}{\mathrm dx}(1+e^x)^{-2}=-2(1+e^x)^{-3}\dfrac{\mathrm d}{\mathrm dx}[1+e^x]=-\dfrac{2e^x}{(1+e^x)^3}$ . Critical points occur for those

where the derivative is 0 or undefined. Neither scenario ever occurs because both the numerator and denominator will be nonzero for any real

.

Because there are no critical points, there will be now local extrema to worry about.

The second derivative is

$\dfrac{\mathrm d}{\mathrm dx}\left[-2e^x(1+e^x)^{-3}\right]=6e^x(1+e^x)^{-4}\dfrac{\mathrm d}{\mathrm dx}[1+e^x]-2e^x(1+e^x)^{-3}$
$=\dfrac{6e^{2x}}{(1+e^x)^4}-\dfrac{2e^x}{(1+e^x)^3}$
$=\dfrac{2e^x}{(1+e^x)^4}\left(3e^x-(1+e^x)\right)$
$=\dfrac{2e^x(2e^x-1)}{(1+e^x)^4}$

Candidates for inflection points are those points where the second derivative vanishes or is undefined (but the original function is still continuous). As before, the denominator is always positive, so the second derivative will always be defined. This time, however, the second derivative will be zero when

$2e^x(2e^x-1)=0\implies 2e^x-1=0\implies e^x=\dfrac12\implies x=\ln\dfrac12\approx-0.6931$

At x=-1

, we have

, and at

, we have

, which means the concavity of f(x)

changes at x=-1

. This means f(x)

is concave downward over $(-\infty,-1)$ and concave upward over $(-1,\infty)$ .

There are no vertical asymptotes to worry about because the denominator is always positive. On the other hand, there are horizontal asymptotes at y=0

and

, as

$\displaystyle\lim_{x\to\infty}f(x)=0$
$\displaystyle\lim_{x\to-\infty}f(x)=1$

Answer from: Quest

i will take them

step-by-step explanation:

Answer from: Quest

a. turk and mcalister

b. howard, mission, and market

c. yes. since streets end some do not cross yet are not parallel to others. example: mcalister and 5th

step-by-step explanation:

parallel lines are lines which do not intersect ever and look a lot like railroad tracks. turk, mcalister and golden gate are examples of these.

intersect are streets which cross each other and a car on one can reach the other. howard, mission, and market intersect with 6th.

skew lines are lines which neither intersect nor are parallel. these lines exist here because some streets end and never intersect another like mcalister and 5th.

A. name the streets that are parallel to golden gate. b. name the streets that intersect 6th street.

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