Question

Let f(x)= (1+e^x)^-2

Answer 1

x-intercepts occur where , and y-intercept occur where .

has no solution because the denominator is always positive for all real values of . Thus there are no x-intercepts.

When , you have , so  intercepts the y-axis at .

The derivative is

. 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

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

At , we have , and at , we have , which means the concavity of  changes at . This means  is concave downward over  and concave upward over .

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

Read more on Brainly.com - brainly.com/question/3642000#readmore

Answer 2

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

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 $f''(-1)$, and at $x=0$, we have $f''(0)>0$, 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 $y=1$, as

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