Question

Calculus:
Find all critical values for f(x) = sqrt(x)*cubert(x-5)

Answer 1

http://www.wolframalpha.com/input/?i=solve+0%3D%28x%5E%281%2F2%29%29%28X-5%29%5E%281%2F3%29

Answer 2

I don't need to find the derivative with the product rule for this one? it's basically how we did the others

Answer 3

yes you do need to use the product rule for this one and then solve it = 0

Answer 4

can we work that part out? I really have trouble with simplifying product rule derivatives when they are composed of roots

Answer 5

I recommend rewriting roots as rational exponents.

Answer 6

\[\sqrt{x}*\sqrt[3]{x-5} \rightarrow x^{1/2}*(x-5)^{1/3}\]

Answer 7

Then "first times the derivative of the second, plus the second times the derivative of the first."

Answer 8

he is correct

Answer 9

It looks a little messy, but isn't too hard to solve.

Answer 10

right again

Answer 11

\[\prime f\]

Answer 12

Do you want to include discontinuities in those critical points, or just extrema?

Answer 13

looks good, but keep it in that top form with the rational exponents - it will make your life a lot easier.

Answer 14

where do i get the critcal values from the equation I wrote down?

Answer 15

Set that equal to zero and solve for x, but first, do you see the discontinuities and domain of x?

Answer 16

Depending on how critical values are defined, they may also include points of discontinuity (holes, vertical asymptotes - where the slope of the tangent line goes to infinity) as well as maxima, minima, and inflection points.

Answer 17

right now we're at only extrama... domain is x =/= 5 and 0. to get the critical values I'd just set the numerator equal to 0 and solve for that?

Answer 18

yep. and you can reduce it to a nice linear equation if you multiply through by the LCM of your denominators to clear your fractions, that will also cancel out all the radicals, and you'll be left with middle school level stuff. ;-)

Answer 19

You should find that there is only one minimum and no maximum.

Answer 20

Find it yet?
\[y'=\frac{1}{3}x^{1/2}\cdot(x-5)^{-2/3}+\frac{1}{2}x^{-1/2}\cdot(x-5)^{1/3}=0.\]
Multiply through by \[6\cdot x^{1/2}\cdot(x-5)^{2/3}\]
\[\rightarrow 2x+3(x-5)=0.\]
Yes?

Answer 21

the next question asks for extrema on interval [0,4] for the same function. now I put in 4 and that yields a nonreal answer in the original function, f(0) and f(5) = 0. is this the sole minimum you were referring to?

Answer 22

You're right about f(0) and f(5), but f(4) is a real number. Now put the critical value of x you found, x* in and compare it to the bounds of the interval. i.e. f(x*)=?