![y=x^5-3\\ y'=5x^4\\\\ 5x^4=0\\ x=0\\ 0\in [-2,1]\\\\ y''=20x^3\\\\ y''(0)=20\cdot0^3=0](https://tex.z-dn.net/?f=y%3Dx%5E5-3%5C%5C%20y%27%3D5x%5E4%5C%5C%5C%5C%205x%5E4%3D0%5C%5C%20x%3D0%5C%5C%200%5Cin%20%5B-2%2C1%5D%5C%5C%5C%5C%20y%27%27%3D20x%5E3%5C%5C%5C%5C%0Ay%27%27%280%29%3D20%5Ccdot0%5E3%3D0)
The value of the second derivative for

is neither positive nor negative, so you can't tell whether this point is a minimum or a maximum. You need to check the values of the first derivative around the point.
But the value of

is always positive for

. That means at

there's neither minimum nor maximum.
The maximum must be then at either of the endpoints of the interval
![[-2,1]](https://tex.z-dn.net/?f=%5B-2%2C1%5D)
.
The function

is increasing in its entire domain, so the maximum value is at the right endpoint of the interval.
Answer:
$30
Step-by-step explanation:
Harry starts out with $62.
The 2 jerseys cost 2($4) = $8.
The ball cost $19.
The shorts cost $5.
The amount Harry has left is:
$62 - ($8 + $19 + $5) = $62 - ($32) = $30
Answer: 
Step-by-step explanation:
1. The formula for calculate th area of a square is shown below:

Where s is the side of the square.
2. Therefore, if the square tabletop has side lengths of (4x-8) units you have:

3. By definition, the square of a binomial is always a trinomial:

4. Then, you obtain:
