91Ó°ÊÓ

To understand the concept of continuity at a specific point, think of a smooth path that has no breaks, jumps, or gaps. For a function to be continuous at a point, the value of the function must be the same from all approaching directions. In our exercise, continuity is examined at the point where the function switches definitions, specifically at \(x = 0\).

Continuity means:

• The left-side limit as \(x\) approaches 0 from the negative direction \((-x)\) is equal to the right-side limit as \(x\) approaches 0 from the positive direction
• These limits must equal the actual value of the function at that point

For the given function \(f(x)=\left\{\begin{array}{cl}e^{a x} & \text { if } \quad x \leq 0 \ 1+\sin (b x) & \text { if } \quad x>0\end{array} .\right.\), we checked and concluded it is continuous at \(x = 0\) because both expressions for the function yield 1 when \(x = 0\).

The left-hand derivative refers to finding the derivative of the function as we approach a particular point from the left side. For our function, we need to consider the portion defined for \(x \leq 0\), which is \(f(x) = e^{ax}\).

To find this derivative at the point \(x = 0\):

• Differentiate the expression \(e^{ax}\) with respect to \(x\), yielding \(f'(x) = a e^{ax}\)
• Substitute into this derivative the value \(x = 0\)
• The result is \(f'(0^-) = a e^{a \cdot 0} = a \cdot 1 = a\)

This process tells us how the function behaves as we approach 0 from the left and provides a rate of change or slope from that direction.

Next, we focus on the right-hand derivative, which considers the change in the function from the right side of the specific point. For our function, we look at the portion for \(x > 0\), expressed as \(f(x) = 1 + \sin(bx)\).

To find this derivative at \(x = 0\):

• Compute the derivative of \(1 + \sin(bx)\), giving \(f'(x) = b\cos(bx)\)
• Substitute \(x = 0\) into this expression
• The answer becomes \(f'(0^+) = b\cos(b \cdot 0) = b \cdot 1 = b\)

This represents the rate of change or slope of the function as it approaches 0 from the right.

For a function to be differentiable at a point, one crucial condition is that the left-hand derivative must equal the right-hand derivative at that point. This means the slope of the function from the left must match the slope from the right, resulting in a smooth transition without any sharp turns or cusps.

In our exercise, this condition is captured by setting \(f'(0^-) = f'(0^+)\). This requirement leads us to the equation \(a = b\). When both derivatives are equal, which happens when \(a\) equals \(b\), it suggests that the function transitions seamlessly through \(x=0\), maintaining the same rate of change from both sides.

Meeting this condition ensures that the function is not only continuous but also differentiable at the point \(x = 0\). This is why, in conclusion, to achieve differentiability at this point, it's necessary for \(a = b\).