Question

Помогите, пожалуйста, с заданием:​

Answer 1

$y=\bigg(\frac{1}{3} \bigg)^{\sqrt{x^2-14x+40}}\\\\\\1. \; x^2-14x+40=0\\\\x^2-4x-10x+40=0\\\\x(x-4)-10(x-4)=0\\\\(x-4)(x-10)=0\\\\x_1=4 \ \ \ x_2=10\\\\\\2. \; x^2-14x+40 \geq 0\\\\x^2-4x-10x+40 \geq 0\\\\x(x-4)-10(x-4) \geq 0\\\\(x-4)(x-10) \geq 0\\\\-\infty\; \;+++4---10+++\; \; +\infty\\\\x \in (-\infty; 4] \; \cup \; [10; +\infty)\\\\\Rightarrow D(f)= (-\infty; 4] \; \cup \; [10; +\infty)$