Answer:
A) f'(c) = 3
Step-by-step explanation:
The mean value theorem says that if f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists some c such that ...
a < c < b
f'(c) = (f(b) -f(a))/(b -a)
__
We are told that f(x) is differentiable on the closed interval [-1, 4], so we know it meets the requirements of the mean value theorem. Then we can conclude that there is some c such that ...
f'(c) = (12 -(-3))/(4 -(-1)) = 15/5
f'(c) = 3 . . . . for some c in the interval -1 < c < 4
Answer:
Step-by-step explanation:
![(2x^3y^4)^{\frac{3}{7}}=\sqrt[7]{(2x^3y^$)^3} =\sqrt[7]{8x^9y^3}](https://tex.z-dn.net/?f=%282x%5E3y%5E4%29%5E%7B%5Cfrac%7B3%7D%7B7%7D%7D%3D%5Csqrt%5B7%5D%7B%282x%5E3y%5E%24%29%5E3%7D%20%3D%5Csqrt%5B7%5D%7B8x%5E9y%5E3%7D)
Answer:
x = 5
Step-by-step explanation:
Step 1: Subtract 2x from both sides.
4x−3−2x=2x+7−2x
<u>2x−3=7</u>
Step 2: Add 3 to both sides.
2x−3+3=7+3
<u>2x=10 </u>
Step 3: Divide both sides by 2.
2x/2 = 10/2
x=5
Answer:
A) 3, 4, 5, 6, 7, 8
Step-by-step explanation:
2+1, 2+2, 2+3, 2+4, 2+5, 2+6= 3, 4, 5, 6, 7, 8
