Understanding the Function y = 3|x| + 1

The function given is \(y = f(x) = 3|x| + 1\). This function involves the absolute value of x, which is defined differently for positive and negative values of x:

• If \(x \ge 0\), then \(|x| = x\), so \(f(x) = 3x + 1\).
• If \(x < 0\), then \(|x| = -x\), so \(f(x) = 3(-x) + 1 = -3x + 1\).

The function can be written piecewise as:

\[f(x) = \begin{cases} 3x + 1 & \text{if } x \ge 0 \\ -3x + 1 & \text{if } x < 0 \end{cases}\]

We are asked to find the Derivative at x=0 for this function.

Calculating the Derivative using Limits

The derivative of a function \(f(x)\) at a point \(x=a\) is defined by the limit:

\[f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\]

or equivalently,

\[f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}\]

To find the Derivative at x=0, we set \(a=0\). First, let's find \(f(0)\):

\[f(0) = 3|0| + 1 = 3(0) + 1 = 1\]

Now, we set up the limit for the derivative at \(x=0\):

\[f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x-0} = \lim_{x \to 0} \frac{(3|x| + 1) - 1}{x} = \lim_{x \to 0} \frac{3|x|}{x}\]

Evaluating Left and Right Hand Limits for Derivative at x=0

For the limit \(\lim_{x \to 0} \frac{3|x|}{x}\) to exist, the left-hand limit and the right-hand limit must be equal.

Right-Hand Limit: As \(x\) approaches \(0\) from the right side (\(x \to 0^+\)), \(x\) is positive, so \(|x| = x\).

\[\lim_{x \to 0^+} \frac{3|x|}{x} = \lim_{x \to 0^+} \frac{3x}{x} = \lim_{x \to 0^+} 3 = 3\]

Left-Hand Limit: As \(x\) approaches \(0\) from the left side (\(x \to 0^-\)), \(x\) is negative, so \(|x| = -x\).

\[\lim_{x \to 0^-} \frac{3|x|}{x} = \lim_{x \to 0^-} \frac{3(-x)}{x} = \lim_{x \to 0^-} -3 = -3\]

Conclusion on Derivative Existence

We found that the right-hand limit of the difference quotient at \(x=0\) is \(3\), and the left-hand limit is \(-3\).

Since the left-hand limit (\(-3\)) is not equal to the right-hand limit (\(3\)), the overall limit \(\lim_{x \to 0} \frac{3|x|}{x}\) does not exist.

Therefore, the Derivative at x=0 for the function \(y = 3|x| + 1\) does not exist. This happens because the graph of the function has a sharp corner at \(x=0\), a characteristic of functions involving the absolute value around the point where the argument of the absolute value is zero.

In calculus, a function must be smooth (no sharp corners or breaks) at a point for its derivative to exist at that point. The Derivative at x=0 signifies the instantaneous rate of change or the slope of the tangent line at that specific point. For functions like \(3|x|+1\), the slope changes abruptly at \(x=0\), preventing a single tangent line or a unique derivative value.