$$f(x)= \begin{cases} x^2 + k & \text{if } x \leq 5 \\ k x\ &\text{if } x > 5 \end{cases} $$
I've seen where you set $x=5$ and plug in, but I don't know how that makes sense. Ive tried $x^2 + k = k x$ and $5^2 + k = k(5)$ so $25 + k = 5k$ but I don't know where to go from here.
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$\begingroup$You can't make it a constant function everywhere as this is a piecewise continous function, where part of it is strongly convex quadratic and another is linear. In particular, $f(1) \neq f(2)$.
From your current working, it seems that you are trying to make it a continuous function, then $$\lim_{x \to 5^-} f(x)=f(5)=\lim_{x \to 5^+} f(x)$$
and what you did is correct. that is you will get $$25+k=5k$$
then $$4k=25$$ and you solve for $k$ by dividing by $4$.
$\endgroup$ 1 $\begingroup$Given function:$$\ f(x) = x^2+k ; (x\le 5)$$$$\ f(x) = kx;(x\gt 5)$$So , a function is continuous at a point($\ a$) if $\ f(a) = \lim_{x\to a^+}f(x) = \lim_{x\to a^-}f(x)$.
Thus , here $\ f$ is continous at $\ x = 5$ if $\ f(5) = \lim_{x\to 5^+}f(x) = \lim_{x\to 5^-}f(x)$.
Now , $\ lim_{x\to 5^+}f(x) = 5k$ and $\ f(5) = \lim_{x\to 5^-}f(x) = 5^2 + k$..
And thus for continuity , $\ 5k = 5^2 + k$ i.e; $\ k = \frac{25}{4}$..
A function $f$ is continuous on an interval if $$ \lim_{x\to a}f(x) = f(a) $$ for every value $a$ in the interval. It is also known that polynomials are continuous everywhere. Your function $$f(x) = \begin{cases} x^2 +k & \text{if } x\leq 5\\ kx & \text{if }5<x \end{cases}$$ will therefore be continuous if $\lim_{x\to 5}f(x) = f(5)$. To accomplish this goal, we can evaluate both polynomials in the piecewise definition of $f$ at $5$ and set them equal: $$5^2 + k = 5k.$$ Solving for $k$ we obtain $k = \frac{25}{4}$ and so $$f(x) = \begin{cases} x^2 +\frac{25}{4} & \text{if } x\leq 5\\ \frac{25}{4}x & \text{if }5<x \end{cases}. $$
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