Question

(9x^2-36x+36)(a-4)/(2^x-a)>=0

Answer 1

$\frac{(9 {x}^{2} - 36x + 36) (a - 4)}{( {2}^{x} - a)} \geqslant 0$
$\frac{9( {x}^{2} - 4x + 4) (a - 4)}{( {2}^{x} - a)} \geqslant 0$
$\frac{{ (x - 2)}^{2} (a - 4)}{( {2}^{x} - a)} \geqslant 0$

(x-2)²всегда ≥0, поэтому имеем:


$\frac{ (a - 4)}{( {2}^{x} - a)} \geqslant 0$
при а≥4
${2}^{x } - a \geqslant 0 \\ {2}^{x} \geqslant a \\ x \geqslant log_{2}(a)$

при 0<а<4
${2}^{x} - a < 0 \\ {2}^{x } < a \\ x < log_{2}(a)$
при а<0
(а-4)<0
(2^х)-а>0
и наше неравенство не имеет решений

Ответ: при а≥4 :
$x \geqslant log_{2}(a) \:$
при 0<а<4 :
$x < log_{2}(a)$

Answer 2

To solve the inequality $(9x^2-36x+36)(a-4)/(2^x-a) \geq 0$, we need to find the intervals for which the expression is non-negative. To do this, we can follow these steps:

Step 1: Identify critical points and discontinuities: The expression is a rational function, so it may have critical points where the numerator or denominator becomes zero and discontinuities where the denominator becomes zero.

Critical points of the numerator (9x^2 - 36x + 36): To find critical points, we set the derivative of the numerator to zero: $\frac{d}{dx} (9x^2 - 36x + 36) = 18x - 36 = 0$ Solving for x: $18x = 36 \Rightarrow x = 2$

Discontinuity of the denominator (2^x - a): The denominator becomes zero when $2^x - a = 0$. Solving for x: $2^x = a \Rightarrow x = \log_2(a)$

Step 2: Create a sign chart: We'll create a sign chart with the critical points and discontinuities we found.

scssCopy code
       x < 2            2 < x < log2(a)       x > log2(a)
(9x^2 - 36x + 36)   -      +        +
(a - 4)             -      -        +
(2^x - a)           +      -        -
Result              -      +        -

Step 3: Analyze the intervals:

For $x < 2$: The numerator $(9x^2 - 36x + 36)$ is positive, and both $(a - 4)$ and $(2^x - a)$ are negative (assuming $a > 4$), so the expression is negative in this interval.

For $2 < x < \log_2(a)$: The numerator $(9x^2 - 36x + 36)$ is positive, and both $(a - 4)$ and $(2^x - a)$ are positive, so the expression is positive in this interval.

For $x > \log_2(a)$: The numerator $(9x^2 - 36x + 36)$ is positive, $(a - 4)$ is negative (assuming $a > 4$), and $(2^x - a)$ is negative, so the expression is negative in this interval.

Step 4: Final solution: The inequality $(9x^2-36x+36)(a-4)/(2^x-a) \geq 0$ holds true for the interval $2 < x < \log_2(a)$.

Please note that the solution assumes $a > 4$ to avoid division by zero at $a - 4 = 0$. Additionally, if $a \leq 4$, the inequality will be satisfied for different intervals or may have a different solution altogether.