Animated Solution for Mathematics - Limits, Continuity and Differentiability: The function $f(x)=(x^2-1)|x^2-3x+2|+\cos (|x|)$ is NOT differentiable at

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dissect a function that, at first glance, looks like a chaotic mess of polynomials and absolute values. We are tasked with finding where $f(x)=(x^2-1)|x^2-3x+2|+\cos (|x|)$ fails to be differentiable.

Many students see the absolute value bars and immediately panic, but I want you to see them as signposts. They are simply telling us where the function might 'bend' or 'break.'

Before we dive into the calculus, let us simplify our environment. We have the term $\cos (|x|)$. Recall that the cosine function is even, meaning $\cos (-\theta )=\cos (\theta )$.

Because of this symmetry, $\cos (|x|)$ is exactly the same as $\cos (x)$. Since $\cos (x)$ is a smooth, oscillating wave that never has a sharp corner, we can safely ignore it in our search for non-differentiability. It is a smooth background, not a source of trouble.

Now, let us look at the polynomial part: $(x^2-1)|x^2-3x+2|$. We should factorize these expressions to see the 'roots' of the problem. We know that $x^2-1=(x-1)(x+1)$ and $x^2-3x+2=(x-1)(x-2)$.

Substituting these back, our function becomes:

Using the property $|ab|=|a||b|$, we can write this as:

Non-differentiability in this context usually occurs at the roots of the absolute value terms, which are $x=1$ and $x=2$. These are the points where the function might experience a sudden change in slope. Let us investigate them one by one.

First, consider $x=1$. Our function near $x=1$ is dominated by the term $(x-1)|x-1|$.

If we look at the behavior of this term, for $x\ge 1$, it is $(x-1{)}^2$, and for $x<1$, it is $-(x-1{)}^2$. If you take the derivative of $(x-1{)}^2$, you get $2(x-1)$, which is $0$ at $x=1$.

Similarly, the derivative of $-(x-1{)}^2$ is $-2(x-1)$, which is also $0$ at $x=1$. Because the left-hand derivative and the right-hand derivative both meet at $0$, the function is perfectly smooth at $x=1$. The 'sharpness' of the absolute value has been 'healed' by the extra factor of $(x-1)$.

Now, let us turn our attention to $x=2$. Near this point, the function behaves like a constant multiplied by $|x-2|$.

Specifically, the coefficient is $(x+1)(x-1)|x-1|$. At $x=2$, this coefficient evaluates to:

So, near $x=2$, our function looks like $3|x-2|+\text{constant}$. Imagine the graph of $3|x-2|$. It is a sharp 'V' shape with a slope of $-3$ on the left and $+3$ on the right.

Because these slopes are not equal, the derivative does not exist at $x=2$. This is our culprit!

We have navigated the function, ignored the smooth cosine, and tested our critical points. We found that while $x=1$ was a false alarm, $x=2$ is a genuine point of non-differentiability.

Remember the golden rule: for a function $g(x)|x-a|$, it is differentiable at $x=a$ if and only if $g(a)=0$. In our case, at $x=2$, the coefficient was $3$, not $0$.

Keep this logic in your toolkit. When you see absolute values, don't fear them—factorize them, check the roots, and look for those sharp corners. You are now ready to tackle any differentiability problem the JEE throws at you!