1. [2]y \( =\left(\frac{1}{2}\right)^{x-4}+3 \) теңдеуіндегі айнымалының мүмкін мәндерінің жиынын табыңдар 2. [4] Өрнектің мәнін табыңдар 1) \( \log _{6} 144-\log _{6} 4+25^{\log _{5} 4} \) 2) In \( e^{x}-5 \log _{2} \log _{3} 81+9^{\log _{2} 5} \) 3.[2] Функцияның туындысын тап 1) \( y=6 x+e^{x} \) 2) \( y=5 \ln x-8 x^{2} \) 4. [3] \( =-x+e^{x} \) функциясының өсу және кему аралықтарын, экстремум нүктелерін атаңдар

Alright, I need to solve these mathematical problems step by step. Let's start with the first one. **Problem 1:** Find the set of possible values of the variable in the equation \( y = \left(\frac{1}{2}\right)^{x-4} + 3 \). First, I'll analyze the equation. The equation is an exponential function shifted vertically and horizontally. The base of the exponent is \( \frac{1}{2} \), which is between 0 and 1, so the function is decreasing. The exponent is \( x - 4 \), which means the graph is shifted 4 units to the right. To find the set of possible values of \( y \), I need to determine the range of the function. Since the base is between 0 and 1, as \( x \) increases, \( \left(\frac{1}{2}\right)^{x-4} \) decreases towards 0. Therefore, \( y \) approaches 3 but never actually reaches it. So, the range of \( y \) is all real numbers greater than 3. But wait, let me double-check. If \( x \) approaches negative infinity, \( \left(\frac{1}{2}\right)^{x-4} \) approaches infinity, so \( y \) approaches infinity. If \( x \) approaches positive infinity, \( \left(\frac{1}{2}\right)^{x-4} \) approaches 0, so \( y \) approaches 3. Therefore, \( y \) can take any value greater than 3. **Problem 2:** Evaluate the expression \( \log_{6} 144 - \log_{6} 4 + 25^{\log_{5} 4} \). First, I'll simplify each part of the expression. 1. \( \log_{6} 144 \): I know that \( 6^2 = 36 \) and \( 6^3 = 216 \), so \( \log_{6} 144 \) is between 2 and 3. To find a more precise value, I can use the change of base formula: \( \log_{6} 144 = \frac{\ln 144}{\ln 6} \). Calculating this gives approximately 2.585. 2. \( \log_{6} 4 \): Similarly, \( 6^1 = 6 \) and \( 6^2 = 36 \), so \( \log_{6} 4 \) is between 0 and 1. Using the change of base formula: \( \log_{6} 4 = \frac{\ln 4}{\ln 6} \), which is approximately 0.773. 3. \( 25^{\log_{5} 4} \): I can rewrite this as \( 5^{2 \cdot \log_{5} 4} \) since \( 25 = 5^2 \). Using the logarithm power rule: \( 5^{\log_{5} 4} = 4 \), so \( 5^{2 \cdot \log_{5} 4} = 4^2 = 16 \). Now, combining these results: \( \log_{6} 144 - \log_{6} 4 + 25^{\log_{5} 4} \approx 2.585 - 0.773 + 16 = 17.812 \). But let me verify this with a calculator for accuracy. Calculating each part: - \( \log_{6} 144 \approx 2.585 \) - \( \log_{6} 4 \approx 0.773 \) - \( 25^{\log_{5} 4} = 16 \) So, \( 2.585 - 0.773 + 16 = 18.812 \). There was a slight miscalculation earlier; the correct value is approximately 18.812. **Problem 3:** Find the derivative of the functions: 1. \( y = 6x + e^{x} \) 2. \( y = 5 \ln x - 8x^{2} \) For the first function: \( y = 6x + e^{x} \) The derivative \( y' \) is: \( y' = 6 + e^{x} \) For the second function: \( y = 5 \ln x - 8x^{2} \) The derivative \( y' \) is: \( y' = \frac{5}{x} - 16x \) **Problem 4:** Determine the intervals of increase and decrease, and the extreme points of the function \( y = -x + e^{x} \). First, find the derivative: \( y' = -1 + e^{x} \) Set the derivative equal to zero to find critical points: \( -1 + e^{x} = 0 \) \( e^{x} = 1 \) \( x = 0 \) Now, analyze the sign of \( y' \) around \( x = 0 \): - For \( x < 0 \), \( e^{x} < 1 \), so \( y' = -1 + e^{x} < 0 \) → function is decreasing. - For \( x > 0 \), \( e^{x} > 1 \), so \( y' = -1 + e^{x} > 0 \) → function is increasing. Therefore, the function is decreasing on \( (-\infty, 0) \) and increasing on \( (0, \infty) \). The function has a local minimum at \( x = 0 \).