Concept:The derivative of f(x)=∣g(x)∣ at a point depends on the sign of g(x). We compute the left-hand and right-hand derivatives separately. If both are equal, the derivative exists and equals that common value.Explanation:Let g(x)=x2−5x+6. Factorize: g(x)=(x−2)(x−3). So g(x)<0 for 2<x<3 and g(x)≥0 elsewhere. Thus f(x)=∣g(x)∣=⎩⎨⎧g(x)−g(x)g(x)x≤22<x≤3x>3. At x=4, we are in the region x>3, so f(x)=x2−5x+6 near x=4. Right-hand derivative at x=4: f′(4+)=h→0+limhf(4+h)−f(4)=h→0+limh(4+h)2−5(4+h)+6−(16−20+6)=h→0+limhh2+8h+16−20−5h+6−2=h→0+limhh2+3h=h→0+lim(h+3)=3. Left-hand derivative at x=4: f′(4−)=h→0+limhf(4)−f(4−h)=h→0+limh2−[(4−h)2−5(4−h)+6]=h→0+limh2−(16+h2−8h−20+5h+6)=h→0+limh−h2+3h=h→0+lim(−h+3)=3. Both left and right derivatives equal 3. Hence f′(4)=3.
