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This problem is an easy application in calculus using the basic ideas of curve sketching. This is the problem 1 from ISI MStat 2019 PSB.

Let \(f(x) = x^{3}-3 x+k,\) where \(k\) is a real number. For what values of

\(k\) will \(f(x)\) have three distinct real roots?

Science is all about asking proper questions. So, \(k\) is the variable here. So, ask the following question,

What role is \(k\) playing here? Let's take \( k = 0\) and see what happens.

\( g(x) = x^{3}- 3x = x(x-\sqrt{3})(x+\sqrt{3})\).

We can easily draw the graph of this \(g(x)\). Let's draw it. So, \(g(x)\) has roots as {\(-\sqrt{3}, 0, \sqrt{3}\)}.

Great, so addding \(k\) to \(g(x)\) will shift the graph upwards right?

Observe that if \(k\) \(\geq\) |lowest value of \(g(x)\) at the bump|, then \(f(x)\) will have less than three roots.

Similarly, observe that if \(k\) \(\leq\) - highest value of \(g(x)\) at the bump, then \(f(x)\) will have less than three roots.

So, let's find them out. So, what is the mathematical significance of these bumps? They are the local maximum and minimum. How do we find them?

\(g'(x) = 0\), where \(g(x) = x^{3}- 3x\).

\( \Rightarrow 3x^2 - 3 = 0 \Rightarrow x = \pm 1\).

So, the minimum and the maximum values of \(g(x)\) are \(g(-1) = 2\) and \(g(1) = -2\).

Hence \( -2 < k < 2\).

Let's see what happens at \( k = \pm 2\).

It is just that point, where the transitition for three roots to two roots occur and then to one root.

Observe that since, it is an odd degree, there will always be a reall root of the curve.

Stay Tuned! Stay Blessed!

- \(f(x)\) has \(n\) distinct roots \(\Rightarrow\) \(f'(x)\) has atleast \(n-1\) distinct roots.
- Is the converse true? i.e. \(f'(x)\) has \(n-1\) distinct roots \(\overset{?}{\Rightarrow}\) \(f(x)\) has atleast \(n\) distinct roots. Can you investigate using the given function?
- \(f'(x)\) has \(n-1\) distinct roots. Does there exist a \(k\) such that \(f(x)+k\) has atleast \(n\) distinct roots?

Stay Tuned! Stay Blessed!

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