f(x)=(1/9)x^9 -a^(8) x. Find the critical point, assuming a is a constant.

Question

anonymous · Answer

The a variable messes me up, not sure how to approach this. Screenshot attached.

campbell_st · Answer

whats the 1st derivative... and whats the 2nd derivative?

anonymous · Answer

err how do i do that with a though?

anonymous · Answer

The first part is x^8, but the next part not sure. Do i need product rule or something?

campbell_st · Answer

ok... 

so 

$$f'(x) = x^8 - a^8$$

$$f"(x) = 8x^7$$

set the 1st derivative to zero and solve

$$0 = x^8 - a^8$$

are you able to solve the derivative..?

anonymous · Answer

Oh i was over complicated it. That makes sense.

anonymous · Answer

Do i solve for x or a though?

anonymous · Answer

$$x=\sqrt[8]{a^8} $$

Is this correct?

campbell_st · Answer

so you have 2 stationary points from the 1st derivative...


use the 2nd derivative to test the nature... 


and solve the 2nd derivative.... for a point of inflection... then check for a change in concavity..

campbell_st · Answer

well you are on the way... what values can x take...?

campbell_st · Answer

if 

$$x = \sqrt[8]{a^8}...then... x = \pm a$$

campbell_st · Answer

I'll leave you to finish... good luck

anonymous · Answer

Right so is that the answer, i don't see what else we could do? x=a?

anonymous · Answer

Seems pointless that x just =a.

anonymous · Answer

or +a, -a

campbell_st · Answer

ok... so you have the critical points or stationary points.. 

there is also 1 in the 2nd derivative 

$$0 = 8x^7$$

but given you don't need to determine there nature... from my point of view the 

critical points are x = -a, x = 0 and x = a 

the values of -a and a are max or mins... x = 0 is a possible point of inflection.

anonymous · Answer

Hmm okay that makes sense i guess.