Given that f (x) = 2x ^ 2-6x + a (a is a constant), there is a maximum value of 3 on [- 2,2]. Find the minimum value of F (x) on [- 2,2]

Given that f (x) = 2x ^ 2-6x + a (a is a constant), there is a maximum value of 3 on [- 2,2]. Find the minimum value of F (x) on [- 2,2]

It is known that f (x) = 2x ^ 2-6x + a (a is a constant), which has a maximum value of 3
There is f (- 2) = 3 = 2 * (- 2) ^ 2-6 * (- 2) + a
The solution is a = - 17
Then f (x) = 2x ^ 2-6x-17
=2(x-3/2)^2-43/2
When x = 3 / 2, the minimum value of function f (x) on [- 2,2] is - 43 / 2

Given that f (x) = 2x ^ 3-6x ^ 2 + m (M has a maximum value of 3 on [- 2,2], then the minimum value of this function on [- 2.2] is

It can be solved by derivative, f '(x) = 6x ^ 2-12x, make it equal to 0, get x = 0 or x = 2
Using the knowledge of derivative, we can know that f ′ (x) has a maximum and x = 2 has a minimum
So now we just need to compare f (- 2) and f (2)
The solution is f (- 2) = - 40 + m, f (2) - 8 + M
Obviously f (x) = at the minimum f (- 2) = - 40 + M

What is the positional relationship between circle x ^ 2 + y ^ 2-4y = 0 and circle x ^ 2 + y ^ 2-16 = 0

Circle x ^ 2 + y ^ 2-4y = 0
x²+(y-2)²=2²
The center of the circle is (0,2), radius = 2
Circle x ^ 2 + y ^ 2-16 = 0
The center of the circle is (0,0); radius = 4
4=2+2
The two circles are inscribed