What value of c in the open interval (0,16)satisfies the Mean Value Theorem for ?
A)0
B)
C)
D)4

Question

​What value of c in the open interval (0,16)satisfies the Mean Value Theorem for  ? ​
A)​0 
B)​ 
C)​ 
D)​4

Quizplus · Accepted Answer

By the Mean Value Theorem, there exists a value c in the open interval (0,16) such that:
 
f'(c)=11eac44f_ed1e_f1e4_bedf_2f595ad74080_TB7497_11  / (16-0)
 
Simplifying this expression gives:
 
f'(c) = 11eac44f_ed1e_f1e4_bedf_2f595ad74080_TB7497_11  / 16
 
To find the value of c that satisfies this equation, we need to take the antiderivative of f(x) and solve for c:
 
f(x) = integral of (3x^2 - 48x + 192) dx = x^3 - 24x^2 + 192x + C
 
where C is the constant of integration. We can solve for C by using the initial condition f(0) = 0:
 
0 = (0)^3 - 24(0)^2 + 192(0) + C
C = 0
 
Thus, f(x) = x^3 - 24x^2 + 192x. Now we can find the value of c:
 
f'(x) = 3x^2 - 48x + 192
 
f'(c) = 3c^2 - 48c + 192
 
3c^2 - 48c + 192 = 11eac44f_ed1f_18f6_bedf_c76379fbc64a_TB7497_11 / 16
 
Multiplying both sides by 16 and simplifying gives:
 
3c^2 - 48c + 192 = 11eac44f_ed1f_18f5_bedf_0d20db00369d_TB7497_11
 
3c^2 - 48c + 192 - 11eac44f_ed1f_18f5_bedf_0d20db00369d_TB7497_11 = 0
 
Using the quadratic formula, we get:
 
c = (-b±sqrt(b^2-4ac))/2a
 
c = (48±sqrt(48^2-4(3)(192-11eac44f_ed1f_18f5_bedf_0d20db00369d_TB7497_11)))/(2(3))
 
c ≈ 11eac44f_ed1f_18f6_bedf_c76379fbc64a_TB7497_11 or c ≈ 4
 
Since c must be in the open interval (0,16), we can eliminate c = 0 and choose c = 11eac44f_ed1f_18f6_bedf_c76379fbc64a_TB7497_11 as the answer.

​What Value of C in the Open Interval (0,16)satisfies the Mean

What Value of C in the Open Interval (0,16)satisfies the Mean