For the first problem, the function will be continuous if $\displaystyle f(2+)=f(2-)=f(2)$ where $\displaystyle 2+$ means some value of $\displaystyle x$ which is just larger than 2 and $\displaystyle 2-$ means some value of $\displaystyle x$ which is just smaller than 2. So obtain the value of $\displaystyle f(2+)=f(2-)$ from $\displaystyle f(x)$ and substitute K by that value. Understood?
if f(2+)= 20
f(2-) = 20
f(2)= k = f(2+) = f(2-)= 20 Is this right ?????
but how can i decide if the function f(x) given in a) have a maxima or a
minima on the intervalls mentioned in the question ?????????????????
Max-Min theorem says :
if f(x) is cont. on the closed interval [a,b], then there exist numbers p,q in the interval [a,b] such that for all x in [a,b]
f(p) less than or equal to f(x) less that or equal to f(q)
Thus f has the absolute min. value m=f(p) taken on at the point p , and the absolute max. value M =f(q)
The function $\displaystyle f(x)=5x^2$ is strictly increasing for $\displaystyle x\geq 0$ so we have:
(i) On $\displaystyle [0,4]$ absolute minimum at $\displaystyle x=0$ and absolute maximum at $\displaystyle x=4$ .
(ii) On $\displaystyle (0,4)$ no absolute minimum and no absolute maximum .
(iii) On $\displaystyle (0,4]$ no absolute minimum and absolute maximum at $\displaystyle x=4$ .
Fernando Revilla
Note, however, that the problem does NOT ask you to determine where the max and min occur, only if they exist.
For (i), at least, it is sufficient to say that the function is continuous on that closed and bounded interval and so both max and min exist.
(ii) and (iii) are a little harder since the max-min theorem does not say what happens if the interval is not closed.
Don't waste a post. Better press the Thanks button.
Fernando Revilla