Question

The values of x at which the real valued function $f(x)=7|2x+1|-19|3x-5|$ is not differentiable is

• A. 1, -1
• B. $\frac{1}{2}, \frac{5}{3}$
• C. $-\frac{1}{2}, \frac{5}{3}$
• D. 0, 1

Hint:

To quickly find the points where a function involving absolute values is not differentiable, set the expression inside each absolute value to zero and solve for $x$. These points are the "critical points" where the function's definition changes, leading to sharp corners in the graph.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The absolute value function, $|u|$, is not differentiable at the point where its argument $u$ is zero. The function has a sharp corner at that point. A function that is a linear combination of absolute value functions, like the one given, will generally not be differentiable at the points where the arguments of any of the absolute value terms become zero.
Step 2: Key Formula or Approach
A function of the form $f(x) = c_1|g_1(x)| + c_2|g_2(x)| + \dots$ is generally not differentiable at the values of $x$ for which $g_i(x)=0$. We need to find the roots of the expressions inside each absolute value.
Step 3: Detailed Explanation
The given function is $f(x) = 7|2x+1| - 19|3x-5|$. This function is a combination of two absolute value functions: 1. The term $|2x+1|$. 2. The term $|3x-5|$. The function $|2x+1|$ has a sharp corner, and is thus not differentiable, at the point where its argument is zero. \[ 2x+1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2} \] The function $|3x-5|$ has a sharp corner, and is thus not differentiable, at the point where its argument is zero. \[ 3x-5 = 0 \implies 3x = 5 \implies x = \frac{5}{3} \] The overall function $f(x)$ is a linear combination of these two functions. The "kinks" from each absolute value function will persist in the combined function (unless a rare cancellation occurs, which is not the case here). Therefore, the function $f(x)$ is not differentiable at these two points. The points of non-differentiability are $x = -\frac{1}{2}$ and $x = \frac{5}{3}$. Step 4: Final Answer
The values of $x$ at which the function is not differentiable are the roots of the arguments of the absolute value functions, which are $-\frac{1}{2}$ and $\frac{5}{3}$.