Find the greatest and least values of \(bx+a\)
for \(-10\leqslant x \leqslant 10\), distinguishing
carefully between the cases \(b>0\), \(b=0\) and \(b<0\).
Find the greatest and least values of \(cx^{2}+bx+a\),
where \(c\ge0\),
for \(-10\leqslant x \leqslant 10\), distinguishing
carefully between the cases that can arise
for different values of \(b\) and \(c\).
Solution:
Case \(b > 0\). Then \(bx+a\) is increasing and the greatest value is \(10b+a\), and the least value \(a-10b\)
Case \(b=0\), then \(a\) is constant and the greatest and least value is \(a\)
Case \(b < 0\), then \(bx+a\) is decreasing and the greatest value is \(-10b+a\) and the least value is \(10b+a\)
If \(c = 0\) we have the same cases as above.
If \( c > 0\) the consider \(2cx+b\). if \(b-20c > 0\) then our function is increasing on our interval and the greatest value is \(100c+10b+a\) and the least value is \(100c-10b+a\)
If \(20c+b < 0\) then our function is decreasing and that calculation is reversed.
If neither of these are true, then the minimum will be when \(x = - \frac{b}{2c}\) and the max at one end point.