Answer:
The minimum value in range of the function is \(f(2)=13\)
Step-by-step explanation:
Be \(y=f(x)=8x-3\) our function to be studied, in the domain \(2\leq x\leq 4\)
If we calculate the derivative \(f'(x)=0\) of \(f(x)\), we can be certain if the function has critical points, where its value could be locally minimum or maximum, so we derive our function
\(f'(x)=8\)
but 8 is different than 0 as we all know, this means that the function doesn't have critical points, this also means that the function is either growing or decreasing in its entire domain, and particularly in the restricted domain of the problem (wich is an interval between 2 and 4, including both values).
So the following step is to plug the interval limit values (2 and 4) into the equation, and by doing this, we get that the minimum value of the range is 13 when the function is evaluated in \(x=2\)
