Limits, Continuity & Differentiability Question - None

Question

If $f(x)=(x^5+1)|x^2-4x-5|+\sin|x|+\cos(|x-1|)$ , then $$f(x)$$ is NOT differentiable at:

Accepted Answer

Step 1: Identify critical points of each absolute value:

$$|x^2-4x-5|=|(x-5)(x+1)|$$ : critical at $$x=-1, 5$$

$\sin|x|$ : critical at $$x=0$$

$\cos(|x-1|)$ : critical at $$x=1$$ (but $\cos|u|$ is even and differentiable at 0)

Step 2: Check x=-1: $$(x^5+1)$$ has factor $$(x+1)$$ which cancels the corner from $$|x+1|$$ . So differentiable at x=-1.

Step 3: Check x=5: $(x^5+1)|_{x=5}=3126 eq 0$ , corner survives. Non-differentiable at x=5.

Step 4: x=0: $\sin|x|$ — LHD=-cos(0)=-1, RHD=cos(0)=1. Non-differentiable.

Step 5: x=1: $\cos(|x-1|)=\cos(|u|)$ which is even and differentiable at u=0. Differentiable.

Total non-differentiable points: x=0 and x=5 → 2 points.

Answer: (A) 2 points

Trap: Missing that (x⁵+1) has (x+1) as a factor, which removes the corner at x=-1.

Key Concept: When |x-a| is multiplied by (x-a), the product is differentiable at x=a

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