\( [-1,5] \)
We are given the function: \[ f(x) = \log_3 (5 + 4x - x^2). \]
Step 1: Find the Domain of the Logarithm
For the logarithmic function to be defined, the argument must be positive: \[ 5 + 4x - x^2>0. \] Rearrange the quadratic inequality: \[ -(x^2 - 4x - 5)>0. \] Factorizing the quadratic expression: \[ -(x-5)(x+1)>0. \] Solving \( (x-5)(x+1)<0 \) using the sign analysis method, the valid interval is: \[ -1<x<5. \]
\Step 2: Find the Range of \( f(x) \)
Since the quadratic expression \( 5 + 4x - x^2 \) is a downward-opening parabola, it attains a maximum value at its vertex. The vertex of the quadratic function \( g(x) = 5 + 4x - x^2 \) is given by: \[ x = \frac{-b}{2a} = \frac{-4}{-2} = 2. \] Substituting \( x = 2 \): \[ g(2) = 5 + 4(2) - (2)^2 = 5 + 8 - 4 = 9. \] Since the logarithm base is 3, we compute: \[ \log_3(9) = 2. \] As \( x \to -1^+ \) or \( x \to 5^- \), the argument \( 5 + 4x - x^2 \) approaches 0, meaning \( \log_3(5+4x-x^2) \) approaches \( -\infty \). Thus, the range of \( f(x) \) is: \[ (-\infty,2]. \]
\[ \text{The domain of the real-valued function } f(x) = \sin^{-1} \left( \log_2 \left( \frac{x^2}{2} \right) \right) \text{ is} \]
Let \( A = [a_{ij}] \) be a \( 3 \times 3 \) matrix with positive integers as its elements. The elements of \( A \) are such that the sum of all the elements of each row is equal to 6, and \( a_{22} = 2 \).
\[ \textbf{If } | \text{Adj} \ A | = x \text{ and } | \text{Adj} \ B | = y, \text{ then } \left( | \text{Adj}(AB) | \right)^{-1} \text{ is } \]
\[ D = \begin{vmatrix} -\frac{bc}{a^2} & \frac{c}{a} & \frac{b}{a} \\ \frac{c}{b} & -\frac{ac}{b^2} & \frac{a}{b} \\ \frac{b}{c} & \frac{a}{c} & -\frac{ab}{c^2} \end{vmatrix} \]