Let f (x) be a function that is differentiable everywhere and has a derivative f'(x) = 3x² + 14x + 6. Verify that the Intermediate Value Theorem for Derivatives applies to the function f' (2) on the interval
[-6, -3], and find the value of c guaranteed by the theorem such that f' (c) = -2

I also attached the work that I've done so far, but I don't think that it's right.

Step-by-step explanation:

F(x)=3x²+14x+6

0=3(-6)²+14(-6)+6

0=108-84+6

0=30

4 and 5)16.97 or 12√2

6)15.59 or 9√3

7)y = 15.59 or 9√3     x =27

8)17.80

the largest number of the four options is 1.8 x 10-2

you have it in scientific notation.

1.80 x 10-6   = 0.00000180

1.80 x 10-4   = 0.000180 (bigger, or "less small", than the former one)

so

1.8 x 10-2       = 0.0180. as you see. the "less small", or the biggest one of them all.

as you have negative exponent, it means you are dividing 1.8 by 100 (i.e. moving the dot towards left, two times).

if you had positive exponent for the scientific notation, power of tenths, you would be multiplying times 100 (moving the dot towards right) as below:

1.8 x (10)2 = 180

so. if you have negative exponent in scientific notation, move the dot toward left. if it is positive, toward right as many times the exponent tells you.