• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q1 (g) • Q1 (h) • Q1 (i) • Q1 (j) • Q1 (k) • Q1 (l) • Q1 (m) • Q1 (n) • Q2 (a) • Q2 (b) • Q2 (c) • Q3 • Q4 (a) • Q4 (b) • Q4 (c) • Q4 (d) • Q4 (e) • Q5 • Q6 • Q7 • Q8 •
Question 04 (d) 

FullSolution Problems. In questions 28, justify your answers and show all your work. If a box is provided, write your final answer there. Simplification of answers is not required unless explicitly requested. Let NOTE: Another notation for tan^{1}(x) is arctan(x) Determine open intervals where the graph of ƒ is concave upwards or concave downwards, and the xcoordinates of all inflection points (if any). 
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! 
Hint 

How does concavity relate to the derivatives of the function f(x)? 
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution 

Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies. To determine the intervals of concavity, we must evaluate ƒ''(x) and determine where it is positive and negative: Evaluating the second derivative of f gives From this we see that the possible inflection points are x = 0. We need only confirm that the sign of the concavity changes across x = 0. We must also consider concavity changes across x = 1 since ƒ is not differentiable there. When x < 1, the second derivative is always negative since 12x^{2} is always negative. When x > 1, the second derivative is always negative as well since in that case, 2x is always negative. From these results we can see that ƒ(x) is concave down for all x except at the critical points. In other words, ƒ(x) is concave down on and so, there are no inflection points. 