This paper discusses the image relationship between the X-Power of function y = A and the - X-Power of y = a (a > 0 and a is not equal to 1), and proves the symmetry about y wrinkle

This paper discusses the image relationship between the X-Power of function y = A and the - X-Power of y = a (a > 0 and a is not equal to 1), and proves the symmetry about y wrinkle

Let's look at it another way:
The x power of function y = 2 and the image of function f (x) are symmetrical about the Y axis, then f (x) is equal to?
On Y-axis symmetry
f(-x)=y=2^x=2^[-(-x)]
f(x)=2^-x

The function f (x) = loga (AX-1) (a > 0 and a ≠ 1) is known (I) find the definition domain of F (x); (II) when x is what value, f (x) > 1?

(1) From the meaning of the question, AX-1 > 0, that is, ax > 1 = A0. When 0 < a < 1, x < 0, that is, the definition domain is (- ∞, 0). When a > 1, x > 0, that is, the definition domain is (0, + ∞); (II) from the meaning of the question, loga (AX-1) > 1 = logaa. When 0 < a < 1, 0 < AX-1 < A, then 1 < ax < A + 1, that is, A0

Let f (x) be a continuous function with t as the period, and ∫ (lower limit a, upper limit x) f (T) DT take t as the period, find ∫ (lower limit 0, upper limit t) f (x) DX =?

Let the original function of F (x) be f (x), ∫ (lower limit a, upper limit x) f (T) DT = f (x) - f (a) = f (x + T) - f (a)
F(x+T)=F(x),F(T)=F(0)
∫ (lower limit 0, upper limit t) f (x) DX = f (T) - f (0) = 0

Y = derivative of (3x-5) to the 3rd / 4th power!

3 / 4 times (3x-5) to the (minus 1 / 4) power and then times 3

Given that the cubic power of function f (x) + the quadratic power of 3x - x + 1 is a subtractive function on R, find the value range of real number a

Missing letter A. if f (x) = ax ³+ 3x ²- X + 1, then f '(x) = 3ax ²+ 6x-1 is constant less than or equal to 0 on R, then: a < 0 and the discriminant is less than or equal to 0, so the range of a can be obtained

If the function f (x) is differentiable on the interval [0, a], and f (a) = 0, it is proved that there is at least one point in the interval (0, a) ξ, Make f( ξ)+ξ f′( ξ)= 0

Construct a new function g (x) = XF (x)
Because g (0) = g (a) = 0
So there must be X
So that G '(x) = 0