If the function f (x) = (A-1) x ^ 2 + ax + 2 is even, what is the monotone increasing interval of function f (x)

If the function f (x) = (A-1) x ^ 2 + ax + 2 is even, what is the monotone increasing interval of function f (x)

F (- x) = f (x), then (A-1) x ^ 2-ax + 2 = (A-1) x ^ 2 + ax + 2
A = 0
So f (x) = - x ^ 2 + 2
Its monotone increasing interval is (negative infinity, 0]
(easy to get from function image)

Given that f (x) = AX2 + BX is an even function defined on [A-1, 2A], then the value of a + B is ()
A. −13B. 13C. −12D. 12

According to the meaning of the title: F (- x) = f (x), B = 0, and A-1 = - 2A, a = 13, a + B = 13

Given that f (x) = AX2 + BX is an even function defined on [A-1, 2A], then the value of a + B is ()
A. −13B. 13C. −12D. 12

According to the meaning of the title: F (- x) = f (x), B = 0, and A-1 = - 2A, a = 13, a + B = 13

A mathematical problem (derivative function)
How to find the derivative of x ^ n

=Nx ^ (n-1) this is the direct application of the formula
deduction
Two proved derivative formulas (e ^ x) '= e ^ x (LNX)' = 1 / X and compound derivative formula are used in the derivation
f(x)=x^n=e^(lnx^n)= e^(nlnx)
f‘(x)= e^(nlnx)*(nlnx)’= x^n*n/x=nx^(n-1)

Do a mathematical problem encountered in the problem, the derivative function
In the stem of the question, "F" (x) | is less than or equal to m, and X is between 0 and a
The answer is:
Let a point C between 0 and a be f (x) maximum point = > F '(c) = 0
How did it come about?
How does f (x) derive such a conclusion? Don't we consider the case that f (x) is not differentiable at the extreme point?
Hope every question can be answered. Thank you

Point C is the maximum value of F (x) between 0 and A. point = > F '(c) = 0
This is because C is the maximum point of F (x) between 0 and a, which is also the maximum of course, and because C is differentiable between 0 and a, f '(c) = 0
This is not derived from the second derivative. It must be differentiable here, because the second derivative is only possible under the condition of first derivative. So it must be differentiable here

Derivative function math problem meeting, please come in
The function f is continuous at I. if the derivative function g of F is discontinuous at a point on I, then G must be the oscillating breakpoint of the second kind of discontinuity at a point. Please talk about it. Don't talk about Darboux theorem. Please be more detailed and sincerely ask for advice
The derivative function is not continuous at a but defined at a

Let f (x) be differentiable in the interval [a, b], then its derivative function can take any value between F (a) and f (b). This is an important feature of the derivative function. Its proof name is as follows: G (x) = f (x) - Rx in [a

For any x ∈ R, the function f (x) = X3 + AX2 + 7ax has no extremum if and only if? Note: X3 is the cubic of X, and AX2 is the square of a times X

The function f (x) = X3 + AX2 + 7ax f '(x) = 3x square + 2aX + 7a the function f (x) = X3 + AX2 + 7ax has no extremum, so f' (x) has no solution, so 4A square - 84A < 0, so a belongs to (0,21), so the function f (x) = X3 + AX2 + 7ax has no extremum if and only if a belongs to (0,21)