II. Verify the Mean-Value Theorem for y = e^x from x = 0 to x =1.

As you can see, the two parts of this problem are irrelevant. The first part of the problem asks for a tangent-line approximation. If we differentiate implicitly for y, we obtain dy/dx to be (y - 35x⁶y⁶ - 18x²y²)/(30x⁷y⁶ - x + 30x³y⁴). The tricky part of this problem is for you to notice that (1,1) is a point on the function. Thus, we can use (1,1) to approximate for (1.05,y). At (1,1) dy/dx is equal to 52/59. By tangent-line approximation,
f(x) = f(x₀) + f'(x₀)*(x - x₀).
Hence, y = 1 + (52/59) * (0.05) = 1.044.
The second part of the problem is the use of Mean Value Theorem. According to this theorem, if f(x) is differentiable from a to b, then there is a "c" where a < c < b, such that f(b) - f(a) = f'(c) * (b-a). For our example, this becomes e¹ - e⁰ = e^c * (1-0). As we can see, c = ln(e-1) which is around 0.541, clearly between 0 and 1.