A derivative question this time: If f''(2) = -6, what is the value of a?

[pinkphiloyd]

I'm still working problems on Khan Academy.

I was just given this problem:

"f(x) = ax^3 + a^2.

If f''(2) = -6, what is the value of a?"

So, I took the first and second derivatives, which, to me, comes out to

f'(x) = 3ax^2 + 2a

f''(x) = 6ax + 2

My final value was incorrect. I eventually looked at the hints, and saw that they were getting

f'(x) = 3ax^2

f''(x) = 6ax


What am I missing/screwing up?

 

[stapel]

Thank you for showing your work and reasoning so nicely! :cool:

"f(x) = ax^3 + a^2.

If f''(2) = -6, what is the value of a?"

So, I took the first and second derivatives, which, to me, comes out to

f'(x) = 3ax^2 + 2a

No. The value represented by "a" is just a constant ("x" is the variable), so the derivative of a² is just zero. Then:

. . . . .\(\displaystyle f(x)\, =\, ax^3\, +\, a^2\)

. . . . .\(\displaystyle f'(x)\, =\, 3ax^2\, +\, 0\, =\, 3ax^2\)

. . . . .\(\displaystyle f"(x)\, =\, 6ax\)

With the second-derivative info, we get:

. . . . .\(\displaystyle f"(2)\, =\, 6a(2)\, =\, 12a\, =\, -6\)

Solve for the value of the constant "a".

 

[Deleted member 4993]

I'm still working problems on Khan Academy.

I was just given this problem:

"f(x) = ax^3 + a^2.

If f''(2) = -6, what is the value of a?"

So, I took the first and second derivatives, which, to me, comes out to

f'(x) = 3ax^2 + 2a

f''(x) = 6ax + 2

My final value was incorrect. I eventually looked at the hints, and saw that they were getting

f'(x) = 3ax^2

f''(x) = 6ax


What am I missing/screwing up?

a² is a constant (not a function of x).

Hence - when you differentiate that w.r.t. 'x' → you get 0.

 

[pinkphiloyd]

Lol. This was one of this problems that I sat and stared at for 30 minutes. As soon as I posted it here, I looked at it again, and it was immediately clear what I had done. Weird how that works!

 

[pinkphiloyd]

Which is not to say that I'm not grateful for the help!