use the mean value theorem to find the c's on the open interval (a,b) such that fprime(c)= (f(b)-f(a))/(b-a)
f(x)= 3xlog(base 2)x , [1,2]
f(x) = 3x log2 x
f(2) = 3(2)(1) = 6
f(1) = 3(1)(0) = 0
so the line joining the endpoints has slope = 6
MVT states there is a c between 1 and 2 such that f'(c) = 6
f'(x) = 3log2 x + 3x/(x ln 2) = 3(log2 x + 1/ln2) = 3/ln2 (lnx + 1)
so, we set f'(c) = 6 and solve for c:
6 = 3/ln2(ln c + 1)
ln c + 1 = 2ln2
ln c = 2ln2 - 1
c = e^(2ln2-1) = 4/e = 1.47
which is in the interval [1,2]
To use the Mean Value Theorem in this case, we need to first verify if the function satisfies the conditions required for the theorem to be applied.
The Mean Value Theorem states that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one number c in the open interval (a, b) such that f'(c) = (f(b) - f(a))/(b - a).
In this case, the function f(x) = 3xlog(base2)x is continuous on the closed interval [1, 2] and differentiable on the open interval (1, 2), since the natural logarithm and exponentiation are differentiable for positive numbers greater than 0.
Now, to find the value of c, we need to calculate f'(x).
To find f'(x), we can apply the product rule and chain rule. Let's start by finding the derivative of 3xlog(base2)x.
Using the product rule, we have:
[f(x)g(x)]' = f'(x)g(x) + f(x)g'(x).
Let f(x) = 3x and g(x) = log(base2)x.
Now, let's calculate the derivatives of f(x) and g(x):
f'(x) = 3,
g'(x) = (1/x) * 1/(ln 2).
Applying the product rule, we get:
[f(x)g(x)]' = f'(x)g(x) + f(x)g'(x)
= 3 * log(base2)x + 3x * (1/x) * 1/(ln 2)
= 3 * log(base2)x + 3/(ln 2).
Now, let's find the value of c.
According to the Mean Value Theorem, there exists at least one number c in the open interval (1, 2) such that f'(c) = (f(2) - f(1))/(2 - 1).
Let's substitute the function and its derivative into the expression:
3 * log(base2)c + 3/(ln 2) = (f(2) - f(1))/(2 - 1).
Evaluate f(2) and f(1):
f(2) = 3 * 2 * log(base2)2 = 6,
f(1) = 3 * 1 * log(base2)1 = 0 (since log(base2)1 = 0).
Substitute the values:
3 * log(base2)c + 3/(ln 2) = (6 - 0)/(2 - 1)
3 * log(base2)c + 3/(ln 2) = 6.
Now, simplify the equation by subtracting 3/(ln 2) from both sides:
3 * log(base2)c = 6 - 3/(ln 2)
log(base2)c = (6 - 3/(ln 2))/3.
To get rid of the logarithm, rewrite the equation in exponential form:
c = 2^[(6 - 3/(ln 2))/3].
Therefore, the value of c in the open interval (1, 2) that satisfies f'(c) = (f(2) - f(1))/(2 - 1) is c = 2^[(6 - 3/(ln 2))/3].