1. (3 points) apply the mean value theorem with a = 2, b = 7 so the function f(x) leads to c = 6. let f(6) =…

Explanation:

Step1: Recall the Mean Value Theorem formula

The Mean Value Theorem states that (f'(c)=\frac{f(b)-f(a)}{b - a}). Here, (a = 2), (b=7), and (c = 6).

Step2: Calculate (f'(6))

Substitute (a = 2), (b = 7) into (\frac{f(b)-f(a)}{b - a}). We don't know (f(2)) and (f(7)), but by the Mean Value Theorem, (f'(6)=\frac{f(7)-f(2)}{7 - 2}). Since (c = 6) satisfies the Mean Value Theorem for (f(x)) on ([2,7]), and the slope of the tangent line at (x = 6) is (f'(6)). Also, (\frac{f(7)-f(2)}{7 - 2}=f'(6)). The slope of the tangent line (m=f'(6)). The equation of a line in point - slope form is (y - y_0=m(x - x_0)). We know (x_0 = 6) and (y_0=f(6)=8). Since (f'(6)=\frac{f(7)-f(2)}{7 - 2}) (by MVT) and the slope of the tangent line at (x = 6) is (f'(6)). The point - slope form of a line is (y-y_1=m(x - x_1)). Here (x_1 = 6), (y_1 = 8) and (m=f'(6)). But from the Mean Value Theorem (m=\frac{f(7)-f(2)}{7 - 2}). However, for the tangent line, we can also think of it in terms of the local linear approximation. The slope of the tangent line at (x = 6) (by the geometric interpretation of the derivative) and using the fact that in the Mean Value Theorem, the slope of the secant line between ((2,f(2))) and ((7,f(7))) is equal to the slope of the tangent line at (x = c=6). The equation of the tangent line using the point - slope form (y - y_0=f'(6)(x - x_0)). Since (x_0 = 6) and (y_0 = 8), and (f'(6)=\frac{f(7)-f(2)}{7 - 2}). But we can also use the fact that the slope of the secant line (\frac{f(7)-f(2)}{7 - 2}) (MVT) is the slope of the tangent line. The point - slope form of a line is (y-8 = f'(6)(x - 6)). And (f'(6)=\frac{f(7)-f(2)}{7-2}). But if we just consider the general form of the tangent line at ((x_0,y_0)=(6,8)) with slope (m) (where (m) is given by the MVT slope of the secant line). The equation of the tangent line is (y-8=\frac{f(7)-f(2)}{5}(x - 6)). Since (c = 6) satisfies (f'(6)=\frac{f(7)-f(2)}{7 - 2}), the equation of the tangent line in point - slope form is (y-8=\frac{f(7)-f(2)}{5}(x - 6)). Simplifying, we can also write it as (y=\frac{f(7)-f(2)}{5}(x - 6)+8). But another way: The Mean Value Theorem gives (f'(6)=\frac{f(7)-f(2)}{7 - 2}). The equation of the tangent line at ((x_0 = 6,y_0=8)) is (y-8=f'(6)(x - 6)). Since (f'(6)) (by MVT) is the slope of the secant line between (x = 2) and (x = 7). Let's assume (m=\frac{f(7)-f(2)}{5}) (from (b - a=7 - 2 = 5)). The point - slope form (y-8=m(x - 6)). Expanding (y=mx-6m + 8).

Answer:

(y-8=\frac{f(7)-f(2)}{5}(x - 6)) (or (y=\frac{f(7)-f(2)}{5}x-\frac{6(f(7)-f(2))}{5}+8))