Working out where a function is non-negative

[srg263]

Hi all,

Any help with part (a) of the below question would be greatly appreciated, and i thank you in advance for your time.

Q) State where the function y= x^4 - 36x^2 is
a) non-negative
b) increasing
c) concave up

Please note i have attached my workings. I think i am confused when saying the sign of the derivative and the function?

View attachment 6671

 

[MarkFL]

Hello and welcome to MHB, srg263! (Wave)

Even though the part of the question you are asking about doesn't involve calculus, since the rest of the question does involve differential calculus, I went ahead and moved this thread to our Calculus forum.

Okay, we are given:

\(\displaystyle y=x^4-36x^2\)

To find where this function is non-negative, we first need to determine where its roots are, so let's factor it completely:

\(\displaystyle y=x^2(x+6)(x-6)\)

Now, we find the roots are:

\(\displaystyle x\in\{-6,0,6\}\)

We should observe that the root $x=0$ is of even multiplicity, which means the sign of the function won't change across this root. The 3 roots divide the real numbers into the following 4 intervals:

\(\displaystyle (-\infty,-6)\)

\(\displaystyle (-6,0)\)

\(\displaystyle (0,6)\)

\(\displaystyle (6,\infty)\)

Now, we want to choose one of these intervals as out test interval, the check the sign of the function within that interval...let's use the 4th interval I listed, and use the value $x=7$...we then find that the signs of the factors are as follows:

\(\displaystyle (+)^2(+)(+) = +\)

Thus, the function is positive in that interval. Now, we can use the fact that the sign will change across roots of odd multiplicity, and won't change across roots of even multiplicity, and that we looking for non-negative values of the function, we must include the roots, can you now state the intervals where the function is non-negative?

 

[srg263]

Hi MarkFL,

Thank you for moving the thread and for taking the time to respond. Your guidance was very helpful. I ended up graphing the function which was also helpful. Am i correct in understanding the function is non-negative (value of the function is >0) in the intervals:
(- ∞, -6) and (6, ∞)

Many thanks.

 

[MarkFL]

Hi MarkFL,

Thank you for moving the thread and for taking the time to respond. Your guidance was very helpful. I ended up graphing the function which was also helpful. Am i correct in understanding the function is non-negative (value of the function is >0) in the intervals:
(- ∞, -6) and (6, ∞)

Many thanks.

Glad to help! :D

You have found where the function is positive, but there's a slight difference between positive and non-negative...positive is greater than zero, but non-negative is greater than or equal to zero, so that's why you need to include the roots...and so this would be:

\(\displaystyle (-\infty,-6]\,\cup\,[0,0]\,\cup\,[6,\infty)\)

 

[srg263]

Ahhhh ha! Yes i can see now i was a bit off by equating non-negative as just positive i.e. >0. Thank you so much! :-)