求参数k的值，使分段函数$f(x)$在任意区间上为常函数

Hey there! Let's walk through this clearly so you get both the "why" and the "how" here.

First: Why do we plug in x=5? #

Your piecewise function is:
$$f(x)= \begin{cases} x^2 + k & \text{if } x \leq 5 \ k x\ &\text{if } x > 5 \end{cases} $$

I assume we're trying to find the value of k that makes this function continuous at x=5 (this is the most common scenario for this type of problem). For a function to be continuous at a point:

• The function must be defined at that point
• The left-hand limit as x approaches the point must equal the right-hand limit
• Both limits must equal the function's value at that point

At x=5:

• The left side (x ≤5) gives us $f(5) = 5^2 + k = 25 + k$
• The right side (x>5) has a limit as x approaches 5 of $k*5 =5k$ (since kx is a linear function, it's continuous everywhere, so the limit equals the value at x=5)

For continuity, these two values have to be equal—that's why we set $x^2 + k = kx$ and plug in x=5.

Next: Solving the equation 25 + k =5k #

You've already got the hard part done by setting up the equation! Now it's just basic algebra:

• Subtract k from both sides to get all the k terms on one side: $25 =5k -k$
• Simplify the right side: $25=4k$
• Divide both sides by 4: $k=25/4$ (or 6.25 if you prefer decimal form)

That's it! Plugging k=25/4 back into the function will make it continuous at x=5.

内容的提问来源于stack exchange，提问作者Emily Burton