Given that any value of even function f (x) on R has derivative, and f '(1) = 1, f (x + 2 = f (X-2), then the slope of tangent of curve y = f (x) at x = - 5 is I use the substitution method to calculate t = 4, f (- 5 + 4) = f (- 1), and then even function f '(- 1) = f' (1) = 1. If I do this, I can calculate k = 1, but the answer is - 1 What's wrong with me in this way or the answer is wrong

Given that any value of even function f (x) on R has derivative, and f '(1) = 1, f (x + 2 = f (X-2), then the slope of tangent of curve y = f (x) at x = - 5 is I use the substitution method to calculate t = 4, f (- 5 + 4) = f (- 1), and then even function f '(- 1) = f' (1) = 1. If I do this, I can calculate k = 1, but the answer is - 1 What's wrong with me in this way or the answer is wrong

T = 4, f (- 5 + 4) = f (- 1), which is true
Even function f '(- 1) = f' (1) = 1, there is a problem here
For even function f (x) = f (- x)
After derivation, we get f '(x) = f' (- x) * (- x) '= - f' (- x)
So f '(- 1) = - f' (1) = - 1
(the derivative of even function is odd function, the derivative of odd function is even function)

The tangent has a slope of - 3 at the inflection point (x * 3)
Y = ax * 3 + BX * 2 + C has an inflection point (1,0), and the tangent slope at this point is - 3. Question: (1) determine the values of parameters a, B and C; (2) determine the monotone interval of the function

1、
y'=3ax²+2bx
y''=6ax+2b
X = 1 is the inflection point
So y '' = 0
6a+2b=0 (1)
The tangent slope is - 3
y'=-3
So 3A + 2B = - 3 (2)
So a = 1, B = - 3
(1,0) on the curve
So 0 = a + B + C
To sum up
a=1,b=-3,c=2
2、
y=x³-3x²+2
y'=3x²-6x=0
x=0,x=2
X 2, y '> 0, y is an increasing function
0

If the tangent equation y = 2x + 1 at point (1, f (x)) on the curve f (x) = x ^ 3 + ax ^ 2 + BX + C, then what is f (x) + F '(x)
```Does anyone know?

It's OK to calculate x = 1 in the lead band. You can calculate it yourself