Question

Writing to Learn Is there a value of \(b\) that will make \(g(x)=\left\\{\begin{array}{l}{x+b} \\ {\cos x}\end{array}\right.\) \(x<0\) \(x \geq 0\) continuous at \(x=0 ?\) differentiable at \(x=0 ?\) Give reasons for your answers.

Short answer

The value of \(b\) that makes the function continuous at \(x = 0\) is \(b = 1\). However, there is no value of \(b\) that makes the function differentiable at \(x = 0\).

Step-by-step solution

Check for Continuity

First, check if the function \(g(x)\) is continuous at \(x=0\). A function is continuous at a point if the limit as \(x\) approaches that point from both sides exist and are equal. Given the piecewise function, compute the limit as \(x\) approaches 0 from the left and from the right, and set them equal to each other: \[\lim_{{x \to 0^-}} (x + b) = \lim_{{x \to 0^+}} (\cos x) \0 + b = 1 \]This gives us \(b = 1\) to make the function continuous.

Check for Differentiability

Now, check if the function \(g(x)\) is differentiable at \(x=0\). A function is differentiable at a point if the limit of the difference quotient exists. This essentially checks if the slope of the tangent line to \(g(x)\) at \(x=0\) matches from both sides. Check this by using the differentiation of each part of the piecewise function and then checking for equality at \(x=0\): \[\lim_{{x \to 0^-}} (1) = \lim_{{x \to 0^+}} (-\sin x) \1 \neq 0 \]No matter what \(b\) is, the two sides of the equation are not equal. Hence, the function is not differentiable at \(x=0\).

Consolidate the findings

From the above steps, we can conclude that \(b = 1\) makes the function continuous at \(x = 0\), but no value of \(b\) can make the function differentiable at that point. Hence, while continuity is achieved, differentiability isn't, and both conditions cannot be satisfied simultaneously.