# Thread: Extrema help

1. ## Extrema help

Hi, after a long break I've decided to look at my math hw since I go back to school tomorrow. Well, I don't understand the first problem so hopefully if I can get help with this one I can understand the rest.

1. (noncalculator) The minimum value of f(x) = x^4 - 4x^3 + k is 7. Determine the value of k

I know how to find min and max values, but this is different so all help is appreciated.

Thanks

2. ## Re: Extrema help

You know that $f_{min}=7=x^4-4x^3+k$ for some fix value of x. What you need to determine is first the value of x when the function has its minimum value, i.e. when $f'(x)=0$. Plug that value of x into the equation above and you'll get k.

3. ## Re: Extrema help

Don't forget to test your critical values obtained from f'(x) = 0. Perhaps the easiest way is to check concavity using f''(x).

4. ## Re: Extrema help

Originally Posted by Ridley
You know that $f_{min}=7=x^4-4x^3+k$ for some fix value of x. What you need to determine is first the value of x when the function has its minimum value, i.e. when $f'(x)=0$. Plug that value of x into the equation above and you'll get k.
Hi, I'm just confused with how I'm supposed to determine the value of x when the function has its min value, for, I have a "k" in the equation which prohibits me from determining x

5. ## Re: Extrema help

k is a constant...what happens to it when you differentiate?

k merely serves to vertically translate the function, does this affect the instantaneous slope?

6. ## Re: Extrema help

Originally Posted by MarkFL2
k is a constant...what happens to it when you differentiate?

k merely serves to vertically translate the function, does this affect the instantaneous slope?
Oh... completely forgot about that, so if k is a constant then the first derivative is simply 4x^3 -12x^2, and when I set that equal to zero I get x=3 and x=0

I now test to see which value of x gives the min by plugging into the equation of second derivative which is: 12x^2 - 24x

when i plug both of these values into the equation of second derivative i get a value of 0 for when i plug in 0 and a value of 36 when i plug in 3, since 36 is positive it indicates concave up which indicates a min, thus x=3 must be my value for x

I then plug 3 into x for original equation and for k I get -20

does this sound correct??

7. ## Re: Extrema help

All of it is good except for finding k.

$f(3)=3^4-4\cdot3^3+k=7$

$3^3\left(3-4\right)+k=7$

$-3^3+k=7$

$k=7+3^3=34$

8. ## Re: Extrema help

Originally Posted by MarkFL2
All of it is good except for finding k.

$f(3)=3^4-4\cdot3^3+k=7$

$3^3\left(3-4\right)+k=7$

$-3^3+k=7$

$k=7+3^3=34$
Oh yeah thanks for catching me on that, I merely got k - 27 = 7 and instead of doing 27+7 I did -27+7, thanks for all the help!