Please solve for all parts
![**Problem 1**: Let \( p \) and \( q \) be the piecewise linear functions given by their respective graphs in Figure 1 below.
[Graph Explanation: Figure 1]
- The graph represents two piecewise linear functions, \( p \) (blue line) and \( q \) (green line).
- The graph is plotted on a Cartesian coordinate system with the x-axis ranging from -3 to 3 and the y-axis ranging from -3 to 3.
- The line \( p \) starts at the point (-3, 3), moves to (-2, 2), then to (-1, 1), followed by (1, 1), (2, -1), and finally to (3, 3).
- The line \( q \) starts at (-3, -3), moves to (0, -2), then to (3, 1).
**(a)** At what values of \( x \) is \( p \) not differentiable? At what values of \( x \) is \( q \) not differentiable? Why?
**(b)** Let \( r(x) = p(x) + 2q(x) \). At what values of \( x \) is \( r \) not differentiable? Why?
**(c)** Determine \( r'(-2) \) and \( r'(0) \).
**(d)** Find an equation for the tangent line and normal line to \( y = r(x) \) at the point where \( x = 2 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F998b6cca-58bb-4391-bb13-94678d180001%2Fabdb772e-4cfd-4320-ac9d-0d5fdd58796e%2Fo1bly1_processed.jpeg&w=3840&q=75)
Part (a)
The function p is not differentiable at x=-1. At this point the left hand limit and the right hand limit are different. Thus, the limit does not exist for this function. That's why, the function is not differentiable.
At x=1 also the function p(x) is not differentiable, because the limit of the function does not exist at this point. The right hand limit and left hand limit shows different values.
So, p is not differentiable at x=-1,1.
The function q is not differentiable at x=-1. At this point the left hand limit and the right hand limit are different. Thus, the limit does not exist for this function. Thatiswhy, the function is not differentiable.
At x=1 also the function q(x) is not differentiable, because the limit of the function does not exist at this point. The right hand limit and left hand limit shows different values.
So, q is not differentiable at x=-1,1.
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