answer:# Part (i): Prove that the sequence ( (a_n) ) is increasing and show that (lim_{n to infty} a_n = 1) when (0 < a_1 < 1). 1. **Show that the sequence is increasing:** Given the recurrence relation ( a_{n+1} = frac{3a_n}{2 + a_n} ), we need to show that ( a_{n+1} > a_n ) for ( 0 < a_n < 1 ). [ a_{n+1} = frac{3a_n}{2 + a_n} ] To show ( a_{n+1} > a_n ): [ frac{3a_n}{2 + a_n} > a_n ] Cross-multiplying: [ 3a_n > a_n(2 + a_n) ] Simplifying: [ 3a_n > 2a_n + a_n^2 ] [ 3a_n - 2a_n > a_n^2 ] [ a_n > a_n^2 ] Since ( 0 < a_n < 1 ), ( a_n > a_n^2 ) holds true. Therefore, ( a_{n+1} > a_n ), and the sequence is increasing. 2. **Show that the sequence is bounded above by 1:** We need to show that ( a_n < 1 ) for all ( n ). Since ( 0 < a_1 < 1 ), assume ( 0 < a_n < 1 ): [ a_{n+1} = frac{3a_n}{2 + a_n} ] Since ( 0 < a_n < 1 ): [ 0 < frac{3a_n}{2 + a_n} < 1 ] Therefore, ( 0 < a_{n+1} < 1 ), and the sequence is bounded above by 1. 3. **Convergence of the sequence:** Since the sequence ( (a_n) ) is increasing and bounded above, by the Monotone Convergence Theorem, it converges to some limit ( l ). 4. **Find the limit:** Let ( lim_{n to infty} a_n = l ). Then: [ l = frac{3l}{2 + l} ] Solving for ( l ): [ l(2 + l) = 3l ] [ 2l + l^2 = 3l ] [ l^2 - l = 0 ] [ l(l - 1) = 0 ] Therefore, ( l = 0 ) or ( l = 1 ). Since ( a_n ) is increasing and ( a_1 > 0 ), ( l neq 0 ). Thus, ( l = 1 ). # Part (ii): Prove that the sequence ( (a_n) ) is decreasing and show that (lim_{n to infty} a_n = 1) when ( a_1 > 1 ). 1. **Show that the sequence is decreasing:** Given the recurrence relation ( a_{n+1} = frac{3a_n}{2 + a_n} ), we need to show that ( a_{n+1} < a_n ) for ( a_n > 1 ). [ a_{n+1} = frac{3a_n}{2 + a_n} ] To show ( a_{n+1} < a_n ): [ frac{3a_n}{2 + a_n} < a_n ] Cross-multiplying: [ 3a_n < a_n(2 + a_n) ] Simplifying: [ 3a_n < 2a_n + a_n^2 ] [ 3a_n - 2a_n < a_n^2 ] [ a_n < a_n^2 ] Since ( a_n > 1 ), ( a_n < a_n^2 ) holds true. Therefore, ( a_{n+1} < a_n ), and the sequence is decreasing. 2. **Show that the sequence is bounded below by 1:** We need to show that ( a_n > 1 ) for all ( n ). Since ( a_1 > 1 ), assume ( a_n > 1 ): [ a_{n+1} = frac{3a_n}{2 + a_n} ] Since ( a_n > 1 ): [ frac{3a_n}{2 + a_n} > 1 ] Therefore, ( a_{n+1} > 1 ), and the sequence is bounded below by 1. 3. **Convergence of the sequence:** Since the sequence ( (a_n) ) is decreasing and bounded below, by the Monotone Convergence Theorem, it converges to some limit ( l ). 4. **Find the limit:** Let ( lim_{n to infty} a_n = l ). Then: [ l = frac{3l}{2 + l} ] Solving for ( l ): [ l(2 + l) = 3l ] [ 2l + l^2 = 3l ] [ l^2 - l = 0 ] [ l(l - 1) = 0 ] Therefore, ( l = 0 ) or ( l = 1 ). Since ( a_n ) is decreasing and ( a_1 > 1 ), ( l neq 0 ). Thus, ( l = 1 ). The final answer is ( boxed{ lim_{n to infty} a_n = 1 } ).

answer:Sarah is grateful for the opportunity to collaborate and is available for a call next week on Tuesday or Thursday after 2 PM. She will compile and send initial findings before the scheduled call to facilitate a productive discussion. Sarah is confident that the collaboration will contribute significantly to eye care education.

answer:Let's start by examining the function f(x) = x^4 - 8x^2 + 16. One immediate observation is that the function contains only even powers of x, which suggests that we can make a substitution to simplify the expression. We can let y = x^2, so that our original function becomes f(y) = y^2 - 8y + 16. This transformation allows us to reduce the degree of the polynomial from 4 to 2. Now, we have a quadratic expression in y, which we can factor as f(y) = (y - 4)^2. Setting this equal to zero, we find that y - 4 = 0, so y = 4. Since y = x^2, we have x^2 = 4, which yields two solutions, x = 2 and x = -2. A clever alternative approach is to notice that the original function f(x) can be expressed as a perfect square, f(x) = (x^2 - 4)^2. From this, it's evident that x^2 - 4 = 0, which gives us the same solutions as before. Both methods lead us to the conclusion that the roots of f(x) are x = 2 and x = -2, each with multiplicity 2. In this problem, a simple substitution and knowledge of quadratic equations allowed us to efficiently find the roots of a quartic polynomial. This example demonstrates the power of clever transformations in solving mathematical problems and the interconnectedness of different areas of mathematics, in this case, the connection between quartic and quadratic equations.

answer:To gain a deeper understanding of the function f(x) = x^4 - 8x^2 + 16 and its roots, let's examine its graph. We already know that the roots of f(x) are x = 2 and x = -2, each with multiplicity 2. We can begin by observing that f(x) is an even function, as f(-x) = f(x). This symmetry property tells us that the graph of f(x) will be symmetric about the y-axis. We can also consider the behavior of f(x) as x approaches positive and negative infinity. Since the leading term x^4 has a positive coefficient, the graph of f(x) will open upwards and eventually become very large as x goes to infinity or negative infinity. Recalling that the function can be expressed as f(x) = (x^2 - 4)^2, we see that f(x) is always non-negative. Therefore, the graph of f(x) will not cross the x-axis except at the points where x^2 - 4 = 0, i.e., at x = 2 and x = -2. When we plot the function, we notice that the graph of f(x) has a double minimum at x = 0, as well as the expected double roots at x = 2 and x = -2. The symmetry of the graph, combined with the knowledge that f(x) is non-negative and the locations of the roots, helps us visualize the overall shape of the graph. Furthermore, we can analyze the relationship between the roots and the graph of the function. We observe that the graph is tangent to the x-axis at the roots x = 2 and x = -2. This indicates that f'(x), the derivative of f(x), must be zero at these points. Since the roots have multiplicity 2, both the function and its derivative vanish at these points. The graph of f(x) provides valuable insight into the relationship between the function's roots and its behavior. The symmetry, asymptotic behavior, and non-negativity of f(x) all contribute to our understanding of the graph and its connections to the roots.