f\left(x\right) = \left({x}^{3} - x\right)\left|\left(x - 1\right)\left(x - 5\right)\right|<br/>= x\left(x + 1\right)\left(x - 1\right)\left|\left(x - 1\right)\left(x - 5\right)\right|<br/>f\left(x\right) = \left\{{(x3−x)(x2−6x+5)−(x3−x)(x2−6x+5)x∈(−∞,1)∪(5,∞)x∈[1,5]\right)
∵(x3−x), \left|{x}^{2} - 6x + 5\right|botharecontinuous\forall x \in R</p><p>∴ f\left(x\right)isalsocontinuous\forall x \in R<br/>Now,<br/>{f}^{'}\left(x\right) = \left\{{(x3−x)(2x−6)+(x2−6x+5)(3x2−1)−(x3−x)(2x−6)−(x2−6x+5)(3x2−1)x∈(−∞,1)∪(5,∞)x∈(1,5)\right)
Now, check differentiability
At x=1
f′(1−)=(13−1)(2−6)+(1−6+5)(3−1)=0
f′(1+)=−(1−1)(2−6)−(1−6+5)(3−1)=0
∵f′(1−)=f(1+)∴atx=1differentiable
At x=5
f′(5−)=(125−5)(10−6)+0
f′(5+)=−(125−5)(10−6)−0
Clearly, f′(5−)=f′(5+)
∴ non-differentiable at x=5
