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Expert: Sherry Wallin - 4/6/2009


Question
I need to know how to find the curve sketching of f(x)=x^4-2x^3+x^2
f(x)=x+1/x+2      f(x)=x-1/x^2

Answer

graph
Hi Olivia~
    Might I assume you have to sketch by hand?
There are a few standard graph techniques that I will share with you.
For the first problem f(x) = x^4-2x^3+x^2 factor it and see where it crosses the x-axis. x^4-2x^3+x^2 = x^2(x^-2x+1) = x^2(x-1)^2 which means that the graph crosses the x axis at 0 and 1. Also you can calculate the first derivative and find the functions 'critical points' these are the places where the derivative is 0 which means it has a local minimum or maximum and if it changes sign then you know it changed it's slope i.e., from neg to pos or pos to neg. So the first derivative is 4x^3-6x^2+2x = 2x(2x^2-3x+1) = 2x(2x-1)(x-1). Set these factors equal to 0. 2x = 0 -> x = 0, 2x-1 = 0 -> x = 1/2, x-1 = 0->x =1
Now break the number line up into intervals with these values of x.
Choose a number in each interval and test whether it's sign is pos or neg. The intervals are (-00,0), (0,1/2), (1/2, 1), and (1, 00)
[I'm using 00 to be infinity]. Now choose a test value in each interval, I will use -1 in (-00,0), 1/4 in (0,1/2), 3/4  in (1/2,1) and 2 in (1,00).  f(-1) = 2(-1)(2(-1)-1)(-1-1) = -2(-3)(-2)= -12. NOTICE that I don't actually need to calculate the exact value BECAUSE all I need is whether each factor is neg or pos. So instead I see that 2(-1) is a neg, (2(-1)-1) is a neg, and (-1-1) is a neg and now I am multiplying a neg*neg*neg an odd number of times which will always result in a neg answer, so to sum this up I know in the interval (-00,0) that x is negative so the slope of the graph in the interval (-00,0) is negative, i.e., going down left to right. I need to check all the other test values f(1/4) = +*-*- = + so the graph turns and goes up in the interval (0,1/2), f(3/4) = = +*+*- = - so in the interval (1/2,1) the graph turns and goes down, and then I check
f(2) = +*+*+ = + so in the interval (1,00) the graph changes again and goes up hill. To check concavity you need to check the second derivative which is f"(x) = 12x^2-12x+2 = 2(6x^2-6x-1). You will need to factor this and use the quadratic formula to get that x ~ .21, .79. Now you need to check a value in the interval (-00,.21) [I used 0],a value in (.21,.79) [I used .5], and a value in (.79,00), [I used 1]. The results I got is that the graph is concave up in (-00,.21)and concave down in (.21,.79), and then concave up in (.79, 00). Note (.21,f"(x)) and (.79, f"(x)) are points of inflection since the original function is continuous at these points and concavity changes.
I will attach a graph that I created with all this information used to graph it.  

Sherry Wallin

Expertise

I can answer most questions up through Calculus and some in Number Theory and Abstract Algebra.

Experience

I have had my Bachelor's Degree since 1987 and have been a teacher since 1988. I earned my Masters Degree in Mathematics May 2010. I have been teaching at the same community college since 2002.

Education/Credentials
I have taught 12 years at the community college level, medical college, and technical college as well as a high school instructor and alternative education instructor and charter school instructor.

Awards and Honors
Master's GPA 3.56 Bachelor's GPA 3.34 Post grad work not degree related GPA 4.0

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