Another Math Problem (Calculus, but more algebra related)

milehigh

I really think Calc would've been cake if I'd had my algebra down better.

I know there's already a math thread going, but I figured a math thread is a bad thread to hijack. I'd hate to be going through responses assuming they were related to my question and then finding that they were actually in response to an altogether different question. Such things may confuse or drive a person insane.

In any case I'm trying to find the inflection points and intervals of concavity for a function. The second derivative of the function (which I've checked to be correct) has the numerator (3x^8-12x^4+1) To get the inflection points I know I need to set this to zero. It's been a long time since I've taken algebra, and for the life of me I can't figure this out.

I know from looking at the graph there's going to be four x intercepts, and hence, 4 places where (3x^8-12x^4+1) will equal zero. This makes me think I need to split this into two smaller equations somehow to get two easier solvable equations, but I get stuck there.

Any hints would be greatly appreciated.

Fizban140

Just a random idea since I am only in college algebra, can you factor that number and set each term to 0?

GoodOmens

I think your best bet would be to make up a dummy variable, something like z=x^4. Substitute that into the equation to make it a quadratic equation with z as teh variable, which should be easy to solve. Once you have the z values, plus them back into your original substitution to find the x values.

That's typically a good way to solve equations like this, with two large exponents once of which is half the other.

iostream

Use the equation (-b+-sqrt(b^2-4ac))/2a when you do the substitution as GoodOmens said. You'll get the answer z=3.914 or z=0.085, which means x=+- of those values since z=x^4 (an even power).

Dance Commander

What's been said is basically correct, but notice that iostream's statement should read
x= +- (those values)^(1/4)

milehigh

Alright...That helped a ton. I'm still getting the hang of substituting stuff. Our teacher used to in working out complex chain rule problems as well (complex for us at least) and my mind is still coming to grips with the whole concept. That makes sense though and I was able to finish the problem. Thanks so much for the help all 8-)

[Michael]

I've always found that using synthetic division on the nastier looking equations can be helpful. Also useful if you're going to be doing curve sketching (without a calculator) to find slant asymptotes.

Terrendos

milehigh wrote: »

Alright...That helped a ton. I'm still getting the hang of substituting stuff. Our teacher used to in working out complex chain rule problems as well (complex for us at least) and my mind is still coming to grips with the whole concept. That makes sense though and I was able to finish the problem. Thanks so much for the help all 8-)

The above method is definitely one of the better ways to go about this. I know this is kind of a cop-out, but I do it all the time in engineering school: if you're allowed, just graph the thing on your calculator. In most of my classes, what's important is that you know how to do the hard stuff, the actual differentials and integrals, since most calculators usually can't help you there. Once you get the algebraic equation, the teachers don't care how you get the final answer.

Keep in mind though, this is after 3 years of Calculus for me, so your mileage may vary with your teachers.