Given the function f (x) = loga (AX-1) (a > 0, a ≠ 1); (1) find the domain of definition of function f (x); (2) discuss the monotonicity of function f (x)
(1) From AX-1 > 0, we get ax > 1. (1 point) when a > 1, X > 0; (2 points) when 0 < a < 1, X < 0. (3 points) so the domain of definition of F (x) is x ∈ (0, + ∞) when a > 1; X ∈ (- ∞, 0) when 0 < a < 1. (4 points) (2) when a > 1, let x1, X2 ∈ (0, + ∞)
RELATED INFORMATIONS
- 1. If y = log2 (x ^ 2 - ax + 1) has the minimum value, then the value range of a is 2 after log is the base
- 2. When the image of the first-order function y = k2x-k + 5 passes through the point (- 2,4), the inverse scale function y = K / X in the quadrant where the image is located, y increases with the increase of X
- 3. If the area of the triangle formed by the image of the first-order function y = KX + 3 and the two coordinate axes is 9, then the value of K is?
- 4. It is known that the image of the first-order function y = KX + B (k > 0) passes through the point P (3,2), and the area of the triangle formed by it and the two coordinate axes is equal to 4. It is better to find several more similar problems in solving the analytic formula of the function~
- 5. If the image of a function y = KX + B passes through points a (0,2), B (1,0), then B+___ ,k=_____
- 6. If the image of a function y = kx-6 passes through a point (- 1,5), then K=______ .
- 7. The image with positive scale function y = KX (K ≠ 0) is a straight line passing through point () and point (1,)
- 8. The image with positive scale function y = KX (K ≠ 0) is processed by___ It's a straight line The image of a linear function y = KX + B (K ≠ 0) is a point (0___ ),(____ , 0)_____ .
- 9. The image of an inverse scale function is shown in the figure. Point a is the upper point on the image, ab ⊥ X axis, AC ⊥ Y axis, and the perpendicular feet are B and C respectively. If the area of the rectangular aboc is 8, then the analytical expression of the inverse scale function is______ .
- 10. As shown in the figure, if point a is on the image of function y = 2x (x > 0), then the area of rectangle aboc is______ A unit of square
- 11. Given the function f (x) = loga (a-ax) (a > 0, and a ≠ 1), find the domain of definition and range of value
- 12. Finding the domain of definition of function y = LG (xsquare-4x-5) Write the process
- 13. F (x) = log2 (x + 1) g (x) = 2log2 (2x + 4) Q to find the domain and minimum value of G-F
- 14. Given that the function f (x) is the domain of definition loga (x + 1) (a > 0, a ≠ 1), the maximum value in the interval [1,7] is 1 / 2 larger than the minimum value, the value of a is obtained
- 15. Find the definition field (1) y = loga (X & # 178; + 1) (a > 0 and a ≠ 1) (2) loga absolute value (x-1) + loga (x + 1) (a > 0 and a ≠ 1) (1) Y = loga (X & # 178; + 1) (a > 0 and a ≠ 1) (2) Absolute value of loga (x-1) + loga (x + 1) (a > 0 and a ≠ 1) (3) F (x) = 1 / radical (1-x) + LG (3 + x)
- 16. Judgment: 1. The function y = a ^ x (a > 0, and a ≠ 1) has the same domain as y = loga (a ^ x) (a > 0, and a ≠ 1), and the function y = 3 ^ (x-1) has the same domain as y = x ^ 3 / X /3. The functions y = 1 / 2 + (1 / (2 ^ x-1)) and y = (1 + 2 ^ x) ^ 2 / (x * 2 ^ x) are odd functions Y = 3 ^ (x-1) has the same range as y = (x ^ 3) / X
- 17. Given that the function f (x) satisfies f (x + 1) = loga (x-1) / (3 + x) (a > 0 and a ≠ 1), the analytic expression and domain of definition of F (x) are obtained
- 18. Given the function f (x) = loga (1-x) + loga (x + 3), where 0
- 19. It is known that the equation loga (x-3) = 1 + loga (x + 2) + loga (x-1) about X has real roots, and the value range of real number a is the bottom I solve it as a > = (7 + 2 √ 10) / 9 or a
- 20. Solve the equation about X: loga (x2-x-2) = loga (X-2 / a) + 1 (a > 0 and a is not equal to 1)