Answer:
f'(0) = -2
f'(-2) = dne
[tex]\lim_{h \to 0^+} \frac{f(2+h)-f(2)}{h} = \text{does not exist}[/tex]
Step-by-step explanation:
The question specifically asks us to use the graph of the function to evaluate the following problems.
In order to find the derivative of a function at a specific point on the graph:
- Choose the point on the graph where you want to find the derivative.
- Figure out the tangent line to this point; this will be the line tangent to the curve of the graph at this point.
- The slope of this tangent line is the derivative at this certain point on the graph.
In order to find f'(0), find x = 0 on the graph. This is at the origin. Now, draw a line tangent to the curve of the graph at x = 0.
This tangent line will follow the same direction as the original function. So, we want to find the slope tangent line now - we can do so by finding the slope of the graph between the interval [-2, 2].
Slope of tangent line for f'(0):
- Slope formula: [tex]\frac{y_2-y_1}{x_2-x_1}[/tex]
- Use coordinate points (-1, 2) and (1, -2)
- Slope of tangent line: [tex]\frac{-2-2}{1-(-1)} = \frac{-4}{2} =-2[/tex]
The slope of the tangent line for f'(0) is -2, therefore, this is the derivative at this point on the graph.
This is solving the problem graphically/algebraically, but you could also look at the graph and determine the rise/run in your head. Your choice.
Now, for f'(-2), we cannot use the same process as before since the graph shows that there is a sharp point at x = -2. This is called a "corner" and means that while the function is continuous, it is not differentiable.
Therefore, we can write that at f'(-2), the derivative does not exist (dne).
The same process occurs for f'(2), since there is a sharp point here. The derivative, based on the graph, does not exist.
