Question

on what intervals is the function f(x)=x^8−8x^7 both decreasing and concave up

Answer 1

Where do the extrema occur (i.e. where are the minimums and maximums)?
$$
\large{
f(x)=x^8−8x^7\\
f'(x)=8x^7-56x^6=0\\
=x^6(8x-56)=0\\
\implies x=0,x=\cfrac{56}{8}=7
}
$$
So they occur when \(x=0\) and \(x=7\).
To check if these are maximums or minimums, take the 2nd derivative:
$$
f''(x)=56x^6-336x^5\\
$$
Which is positive when \(56x^6-336x^5>0\) or \(56x^5((x-6)>0\) or \(x>6\). So when \(x>6\), the function, \(f(x)\) is increasing and when it is less than this, it is decreasing. Therefore, \(f(7)\) is a minimum, since \(x=7>6\). Similarly, \(f(0)\) is a maximum, since \(x=0<6\).

But we need to know the interval where the function is decreasing and concave up. We know that \(f(x)\) is decreasing when \(0<x<7\) and is concave up when \(f''(x) > 0\), which occurs when \(x> 6\).

\(\qquad \bf \text{Therefore, the interval we seek is \(6 < x < 7\)}\).

Note that f(0) is really a saddle point and not a real maximum and f(7) is a global minimum with value \(f(7)=-823,543\)

Here is a chart that confirms some of these observations:
http://www.wolframalpha.com/input/?i=x^8%E2%88%928x^7+

Answer 2

*I should have said that the second derivative at f(0) was inconclusive (because f''(0)=0), and so could not be a maximum. But we know that at \(0<x<7\), f(x) is decreasing because \(f''(x)<0\) there, which is the information we need for this problem.