In this case, the slope of the tangent line at the point (2, 11/2) is 112.
The given question seems to be asking for the slope of the tangent line to the function f(x) = 8(2x-3)^7 at the point (2, 11/2).
To find the slope of tangent line, we can use the derivative of the function. The derivative of f(x) with respect to x gives us the rate of change of the function at any given point.
First, let's find the derivative of f(x). To do this, we can use the chain rule. The chain rule states that if we have a function g(x) raised to a power n, then the derivative of g(x)^n with respect to x is n * g(x)^(n-1) * g'(x), where g'(x) is the derivative of g(x).
Using the chain rule, the derivative of f(x) = 8(2x-3)^7 with respect to x is:
f'(x) = 8 * 7 * (2x-3)^(7-1) * (d/dx) (2x-3)
Simplifying this, we have:
f'(x) = 56 * (2x-3)^6 * 2
Now, let's evaluate the derivative at x = 2 to find the slope of the tangent line at the point (2, 11/2):
f'(2) = 56 * (2*2-3)^6 * 2
= 56 * (4-3)^6 * 2
= 56 * (1)^6 * 2
= 56 * 1 * 2
= 112
Therefore, the slope of the tangent line to the function f(x) = 8(2x-3)^7 at the point (2, 11/2) is 112.
In summary, to find the slope of the tangent line, we first find the derivative of the function with respect to x using the chain rule. Then, we evaluate the derivative at the given point to find the slope. In this case, the slope of the tangent line at the point (2, 11/2) is 112.