25. find the interval(s) over which the function is continuous. you may write the answer using interval…

Explanation:

Step1: Analyze function a

For $y = \sqrt{8x - 16}$, the expression under the square - root must be non - negative. $8x-16\geq0$. Solve the inequality: $8x\geq16$, so $x\geq2$. In interval notation, the interval of continuity is $[2,\infty)$.

Step2: Analyze function b

For $y=\ln(2 - x)$, the argument of the natural logarithm must be positive. $2 - x>0$. Solve the inequality: $-x>-2$, so $x < 2$. In interval notation, the interval of continuity is $(-\infty,2)$.

Step3: Analyze function c

The exponential function $y = e^{4x - 1}$ is continuous for all real numbers. The domain of $y = e^{4x - 1}$ is $(-\infty,\infty)$.

Step4: Analyze function d

For $y=\frac{4x}{x^{2}-4}$, the denominator cannot be zero. Set $x^{2}-4=(x + 2)(x - 2)=0$. The roots are $x=-2$ and $x = 2$. The function is continuous for all real numbers except $x=-2$ and $x = 2$. In interval notation, the intervals of continuity are $(-\infty,-2)\cup(-2,2)\cup(2,\infty)$.

Step5: Analyze function e

The cosine function $y=\cos x$ is continuous for all real numbers. The domain of $y = \cos x$ is $(-\infty,\infty)$.

Answer:

a. $[2,\infty)$ b. $(-\infty,2)$ c. $(-\infty,\infty)$ d. $(-\infty,-2)\cup(-2,2)\cup(2,\infty)$ e. $(-\infty,\infty)$