Problem 52 - College Algebra and Calculus: An Applied Approach

Question

Describe the \(x\) -values at which the function is differentiable. Explain your reasoning. $$ y=\left|x^{2}-9\right| $$

Step-by-Step Solution

Answer

The function \(y=\left|x^{2}-9\right|\) is differentiable at all \(x\)-values except \(x=-3\) and \(x=3\).

Step 1: Establish the Inside Function

Firstly, let's focus on the function inside the absolute value, which is \(x^2 - 9\). This function is a quadratic and thus, it is differentiable for all real numbers.

Step 2: Determine Critical Points

Next, we need to check the value of x for which \(x^2 - 9\) is 0. Setting \(x^2 - 9 = 0\), we get that \(x= -3\) or \(x=3\). These two points split the function into three intervals \((- \infty, -3)\), \((-3, 3)\) and \((3, \infty)\).

Step 3: Define the Function for Different Intervals

The absolute value function is defined as \(|a| = a\) if \(a ≥ 0\) and \(|a| = -a\) if \(a < 0\). Therefore, the function \(|x^2 - 9|\) can be redefined for different intervals as follows: On the interval \((- \infty, -3)\), \(x^2 - 9 < 0\), so \(|x^2 - 9| = -(x^2 - 9)\). On the interval \((-3, 3)\), \(x^2 - 9 ≥ 0\), so \(|x^2 - 9| = x^2 - 9\). On the interval \((3, \infty)\), \(x^2 - 9 > 0\), so \(|x^2 - 9| = x^2 - 9\). Each of these piece-wise functions is differentiable within its interval.

Step 4: Examine The Differentiability at Critical Points

Finally, we need to check for differentiability at critical points \(x=-3\) and \(x=3\). By examining the function \(y=|x^2 - 9|\) at these points, it's clear that the function does not have a defined derivative at \(x=-3\) and \(x=3\) because the absolute value function makes a sharp turn at these points. Therefore, function \(y=|x^2-9|\) is differentiable at all points except \(x=-3\) and \(x=3\).