If the tangent of the curve y = X3 + X-2 at point P0 is parallel to the straight line y = 4x, then the coordinate of point P0 is () A. (0, 1) B. (1, 0) C. (- 1, - 4) or (1, 0) d. (- 1, - 4)

If the tangent of the curve y = X3 + X-2 at point P0 is parallel to the straight line y = 4x, then the coordinate of point P0 is () A. (0, 1) B. (1, 0) C. (- 1, - 4) or (1, 0) d. (- 1, - 4)

Let the coordinates of point P0 be (a, f (a)), f '(x) = 3x2 + 1 is obtained from F (x) = X3 + X-2, and the tangent of the curve at point P0 is parallel to the straight line y = 4x. The slope of the tangent equation is 4, that is, f' (a) = 3a2 + 1 = 4. The solution is a = 1 or a = - 1. When a = 1, f (1) = 0; when a = - 1, f (- 1) = - 4, then the coordinates of point P0 are (1, 0) or (- 1, - 4)

It is known that the function f (x) = x ^ 2 + (M + 1) x + 2m is even, and the tangent equation f (x) at x = 1 (n-2) x-y-3 = 0
Ze, then the product of the constant Mn is equal to zero

Since f (x) is an even function, f (- x) = f (x), that is, x ^ 2 - (M + 1) + 2m = x ^ 2 + (M + 1) x + 2m
The solution is m = - 1 and f (x) = x ^ 2-2;
So f '(x) = 2x tangent slope k = f' (1) = 2, so n-2 = 2, n = 4
mn=-4

Let f (x) be an even function. If the tangent slope of the curve y = f (x) at point (|, f (|) is |, then the tangent slope of the curve at point (1, f (1)) is |?

(x) It's an even function
f(-x)=f(x)
So: F '(- x) * (- x)' = f '(x), f' (- x) = - f '(x)
f(-1)=f(1)
f'(-1)=-f'(1)=-1
Then the slope of the tangent at (- 1, f (- 1)) is - 1