The function f is continuous on the interval [3, 13] with selected values of x and f(x) given in the table below. Use the data in the table to approximate f '(3.5)

x 3 4 7 10 13
f(x) 2 8 10 12 22

I think what you're supposed to do is find the point between three and four since it is f(3.5).

This means you would do slope formula, which is (8-2)/(4-3), and you would get 6.

Yes you are correct - 6 is the answer!

Well, isn't this table looking quite enigmatic? It's like a math comedy show but without the jokes! Alright, let's get to solving this.

To approximate f'(3.5), we'll use the concept of average rate of change. We can find the average rate of change between 3 and 4, and see if it gives us a rough idea of the derivative at 3.5.

So, the average rate of change between x = 3 and x = 4 is given by:

(f(4) - f(3))/(4 - 3)

Substituting the values from the table, we have:

(8 - 2)/(4 - 3)

This boils down to:

6/1

So, our approximate value of f'(3.5) is 6.

Keep in mind that this is just an approximation. For a more accurate calculation, you would need more data points or additional information. But hey, math jokes aside, I hope this helps!

To approximate f'(3.5) using the data in the table, we can use the concept of finite differences.

Step 1: Calculate the differences in the values of f(x) for adjacent x-values:
Δf(x) = f(x+1) - f(x)

Using the given values in the table, we have:
Δf(3) = f(4) - f(3) = 8 - 2 = 6
Δf(4) = f(7) - f(4) = 10 - 8 = 2
Δf(7) = f(10) - f(7) = 12 - 10 = 2
Δf(10) = f(13) - f(10) = 22 - 12 = 10

Step 2: Calculate the differences in the x-values:
Δx = x+1 - x = 1

Step 3: Calculate the finite difference quotient:
f'(3.5) ≈ Δf / Δx = Δf(3) / Δx

Substituting the values calculated from step 1 and step 2:
f'(3.5) ≈ 6 / 1 = 6

Therefore, f'(3.5) is approximately 6.

The function is clearly not linear, so I'd just go with the slope of the segment from f(3) to f(4).

That would make f(3.5) = (2+8)/2 = 5

Plotting the points makes the graph look like a cubic with the greatest curvature near (7,10) so our approximation should be fairly close.