Find the value of $g \left(\frac{1}{4} \right)+g \left(\frac{1}{2} \right)+g \left(\frac{5}{4} \right)$ If $$f(x)=4x^3-x^2-2x+1$$ and $$g(x)=\begin{cases} 
      \min\{f(t):0 \leq t\leq x\} & 0\leq x\leq 1\\
      3-x & 1\leq x\leq 2 
   \end{cases}$$
then find the value of $$g \left(\frac{1}{4} \right)+g \left(\frac{1}{2} \right)+g \left(\frac{5}{4} \right)$$
Could someone explain to me what does the condition $\min\{f(t):0 \leq t\leq x\}$, $  0\leq x\leq 1$ mean and how to use it here?
 A: I suspect that you’ve left off ‘$g(x)=$’ before the displayed cases. That is, I suspect that it should read 
$$g(x)=\begin{cases}
\min\{f(t):0\le t\le x\},&\text{if }0\le x\le 1\\
3-x,&\text{if }1\le x\le 2\;.
\end{cases}$$
If so, it means simply that (for example) 
$$g\left(\frac14\right)=\min\left\{f(t):0\le t\le\frac14\right\}\;,$$
the minimum value of $f(t)$ on the interval $\left[0,\frac14\right]$. This is found in the usual way: differentiate $f$ and determine its minimum value on this interval.
A: I am going to take a guess and assume the problem actually read:
"$g(x) = \begin{cases} 
      \min\{f(t):0 \leq t\leq x\} & 0\leq x\leq 1\\
      3-x & 1\leq x\leq 2 
   \end{cases}$"
the phrase:
"\begin{cases} 
      \min\{f(t):0 \leq t\leq x\} & 0\leq x\leq 1\\
      3-x & 1\leq x\leq 2 
   \end{cases}"
By itself just doesn't make sense. (It's like a sentence without a verb.)
""$g(x) = \begin{cases} 
      \min\{f(t):0 \leq t\leq x\} & 0\leq x\leq 1\\
      3-x & 1\leq x\leq 2 
   \end{cases}$"
means that $g$ is a function that is defined to be: $g(x) = 3- x$ whenever $x \in [1,2]$.  However if $x \in [0,1]$ then $g(x)$ is define to be $\min\{f(t):0 \leq t\leq x\}$.  $\min\{f(t): 0 \le t \le x\}$ means precisely what it seems:  The minimum value of $f(t)$ for a specific range of $t$.
In your problem as $ 1/2 < 1$ you use $g(1/2) = \min\{f(t):0 \leq t\leq 1/2\}$ but as $5/4 > 1$ you use $g(5/4) = 3 - 5/4$.
