Calc BC problems. Need work

f'(0) = -2

f'(-2) = dne

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:

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):

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.

you have 6 equal triangles, find the area of one of them (b*h/2) and multiply by 6

(8*4: 2)*6 =

32 cm² is your answer