91Ó°ÊÓ

Continuity is a fundamental concept in calculus. It refers to the smoothness of a function at a point. If a function is continuous at some point, the function does not "jump" or have any "holes" at that point.

To check continuity at a particular point, like at \( x = 1 \) in our exercise, we must ensure three things:

• The left-hand limit as \( x \to 1^- \) (approaching from the left) exists.
• The right-hand limit as \( x \to 1^+ \) (approaching from the right) exists.
• Both these limits are equal to the function's value at that point, \( f(1) \).

In our task, continuity at \( x = 1 \) requires that \( \lim_{{x \to 1^-}} f(x) = f(1) = \lim_{{x \to 1^+}} f(x) \). Since we calculated \( \lim_{{x \to 1^-}} f(x) = -1 \), the function value \( f(1) \) must also be \( -1 \). Thus, \( a + \cos^{-1}(1+b) = -1 \). This condition ensures continuity of the function at the specified point.

The left-hand derivative of a function measures the derivative as you approach a point from the left side. It's crucial for determining if a function is smooth at a boundary point, such as \( x = 1 \).

Our function \( f(x) = -x \) for \( x < 1 \), provides a straightforward calculation. The derivative of \( f(x) = -x \) is simply \( -1 \).

Therefore, the left-hand derivative of \( f \) at \( x = 1 \) is \(-1\). This derivative shows us that from the left side, the function descends at a constant rate of \(-1\).
In essence, ensuring the left-hand derivative exists and equals the right-hand derivative is part of proving differentiability at a point.

The right-hand derivative is the derivative as you approach from the right of a certain point, like \( x = 1 \) in our example. This derivative is equally important as the left-hand derivative in determining the function's differentiability.

For \( x \geq 1 \), the function given is \( a + \cos^{-1}(x+b) \). We need to differentiate this expression with respect to \( x \).

The chain rule helps us here. Differentiating \( \cos^{-1}(x+b) \) using chain rule, we get \(-\frac{1}{\sqrt{1-(x+b)^2}}\). So originally with \( b=0 \), we find \(-\frac{1}{\sqrt{1-1^2}} = -\infty \), which was invalid for differentiability. This meant recalibrating \( b \) to \( -1 \) to satisfy derivative criteria, resulting in equality at \(-1\).
Aligning this so that both left and right derivatives at \( x = 1 \) are equal is key to establishing differentiability. It's why further corrections proving \( b = -1 \), along with \( a = -1 \) work and equalize both derivatives to \(-1\).

The Chain Rule is a powerful tool in calculus for differentiating compositions of functions. It allows us to handle more complex expressions by breaking them into simpler parts to differentiate.

When faced with a situation where you have \( \cos^{-1}(x+b) \), recognizing it as a composition of functions is essential. The outer function \( y = \cos^{-1}(u) \) and the inner function \( u = (x+b) \). The chain rule states that the derivative of a composition \( y(u(x)) \) is:

• The derivative of the outer function with respect to its argument \( \frac{dy}{du} \),
• multiplied by the derivative of the inner function with respect to \( x \), \( \frac{du}{dx} \).

Applying this process, the derivative of \( y = \cos^{-1}(x+b) \) is \(-\frac{1}{\sqrt{1-(x+b)^2}}\). This technique was pivotal in our problem when analyzing differentiability from the right at \( x = 1 \). Using the chain rule, we ensured the differentiation aligned correctly, which helped in solving for parameters like \( b = -1 \).