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.

80x^8−68x^6+16^5−20x^4−4x^3+8x^2

hope i'm right lol

step by step resolutions

1) n+n+1+n+2=51

3n+3=51

3n=48

n=16

the three numbers are 16, 17 and 18.

2) x+15=4x

3x=15

x=5

radha is now 5 years old

3) the two digits are x and y.

y=3x

(10x+y)+(10y+x)=88

y=3x

11x+11y=88

y=3x

x+y=8

x+3x=8

x=2

the original number can be either 28 or 82.

4) n+(n+2)+(n+4)=48

3n+6=48

3n=42

n=14

the numbers are 14, 16 and 18.

5) y=x+3

10y+x+10x+y=143

y=x+3

11x+11y=143

y=x+3

x+y=13

x+x+3=13

2x=10

x=5

the original number can be 57 or 75.

hope this !