![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.
We solve the equation, ( a + a + 1 )^2 = 112 + a^2 + ( a + 1 )^2;
Then, ( 2a + 1 )^2 = 112 + a^2 + a^2 + 2a +1;
4a^2 + 4a + 1 = 113 + 2a^2 + 2a;
Finally, 2a^2 + 2a - 112 = 0;
a^2 + a - 56 = 0;
We use <span>Quadratic Formula for this Quadratic Equation;
The solutions are a1 = 7 and a2 = -8;
But a is a natural number; so, a = 7;
The natural consecutive numbers are 7 and 8.</span>
Answer:
10
Step-by-step explanation:
Answer:
35 multiplied by 12
Step-by-step explanation:
Because either you multiply it or you do sohcahtoa
