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When we talk about the continuity of a function, we mean that the function behaves nicely without any jumps or breaks. In mathematical terms, a function \( f(x) \) is continuous on an interval \([a, b]\) if it can be drawn on a graph without lifting the pencil. For Rolle's Theorem to be applicable, the function must be continuous on the closed interval \([a, b]\). The function given in the original exercise is \( f(x) = (x-1)\sin x \). This is a combination of a linear factor \( (x-1) \) and a trigonometric function \( \sin x \). Linear functions and the sine function are both well-known examples of continuous functions. Therefore, \( f(x) \), being a product of these elementary functions, is continuous on the interval \([0, 1]\). When solving problems using Rolle's Theorem, always remember:

• The function must be defined and continuous on the entire closed interval.
• This ensures there are no gaps or holes in the graph of the function within that interval.

Differentiability is all about whether a function has a derivative at every point in a particular interval. A function \( f(x) \) is said to be differentiable on an interval \((a, b)\) if it has a defined derivative \( f'(x) \) at every point within that interval. In simple terms, if you can find the slope of the tangent line to the function at any point in \((a, b)\), the function is differentiable there. For Rolle's Theorem to apply, the function must be differentiable on the open interval \((a, b)\). With \( f(x) = (x-1)\sin x \) in the problem, since both \( x-1 \) and \( \sin x \) are differentiable functions, their product is also differentiable on \((0, 1)\). Here's a quick checklist to confirm differentiability:

• The function should be smooth without any sharp corners or cusps.
• The derivative should be defined for every point in the open interval.

Remember that continuity does not always imply differentiability. But if a function is differentiable, it is also continuous.

The product rule is a fundamental technique in calculus used to determine the derivative of the product of two functions. If you have a function \( h(x) = u(x)v(x) \), where \( u(x) \) and \( v(x) \) are both differentiable, then the derivative \( h'(x) \) is given by:\[ h'(x) = u'(x)v(x) + u(x)v'(x) \] In the given exercise, the function to differentiate using the product rule is \( f(x) = (x-1)\sin x \). Here's how we do it:

• Let \( u(x) = (x-1) \) so \( u'(x) = 1 \).
• Let \( v(x) = \sin x \) so \( v'(x) = \cos x \).
• Apply the product rule: \[ f'(x) = (x-1)\cos x + \sin x \]

Understanding the product rule is crucial when dealing with functions where two variables are multiplied together. It's a versatile tool that makes tackling a variety of calculus problems much simpler.

Trigonometric functions are a core part of mathematics, especially in calculus. They are functions of angles and are very useful in waves, oscillations, and periodic phenomena.The most common trigonometric functions are sine, cosine, and tangent, denoted as \( \sin(x) \), \( \cos(x) \), and \( \tan(x) \), respectively. In the exercise, we focus on \( \sin(x) \) and \( \cos(x) \), especially in the context of differentiating the function \( f(x) = (x-1)\sin x \).Key things to remember about trigonometric functions:

• \( \sin(x) \) and \( \cos(x) \) are periodic, meaning they repeat their values in a regular interval.
• They are continuous and differentiable everywhere, except at certain points for \( \tan(x) \) which is not involved directly in differentiability here.
• The derivative of \( \sin(x) \) is \( \cos(x) \).

Being familiar with trigonometric functions and their properties helps in understanding many calculus concepts, like applying Rolle's Theorem as seen here.