I have the following question here.

a) Let $f(x)=x^4-x^3+1$. Show that the graph of the function $f$ lies above the $x$-axis

Is there a way to approach this in a "nice" way? I could just sketch the graph using derivatives, concavity and such but is there a more nicer way of doing this? I feel like I could do this using integration in some way but I am not really sure.

b) Consider the region bounded by the graph of $f$, the $x$-axis, the line $x = a$, and the line $x = a + 1$. What is the value of a for which the area of this region reaches its minimum? What is the value of this minimum?

Would I simply take the bounds as $a$ and $a+1$ and then integrate the function in terms of $a$, and then maximize the function by taking the derivative with respect to $a$?

Any help would be very much appreciated!


Try to express it as a sum of squares. One way that I did it was

$$ x^4 - x^3 + 1 = (x^2 - \frac{1}{2} x - \frac{1}{2} ) ^2 + \frac{1}{12} (3x-1)^2 + \frac{2}{3}.$$

Another way is

$$ x^4 - x^3 + 1 = (x^2 - \frac{1}{2} x - \frac{1}{4} ) ^2 + \frac{1}{16} (2x-1)^2 + \frac{7}{8}.$$

  • $\begingroup$ nice(+1) but how did you come up with that,i am curious $\endgroup$ – Albus Dumbledore Nov 17 '20 at 5:42
  • $\begingroup$ @AlbusDumbledore Given how much leeway there is with the bound, there are lots of "nice" SOS we could try. Wishful thinking that the quadratic square should look like $x^2 - \frac{1}{2}x + C$, so just play with that. We're then left with a quadratic, which we just need to test with the discriminant. $\endgroup$ – Calvin Lin Nov 17 '20 at 5:44
  • $\begingroup$ i see ,thanks for insight am-gm also helps as i did below $\endgroup$ – Albus Dumbledore Nov 17 '20 at 5:46
  • $\begingroup$ This was my original plan but I didn't know how to go about doing this. Thank you! $\endgroup$ – Future Math person Nov 17 '20 at 6:06

Hint: take cases when $x\le 0$ , $x\in[0,1]$ and $x\ge 1$

$x\le 0$:$x^4+1+\underbrace{(-x^3)}_{\ge 0}> 0$

$x\in [0,1]$:$x^4+\underbrace{(1-x^3)}_{\ge 0}>0$

for $x\ge 1$ its obvious

  • $\begingroup$ +1 (Though you might want to elaborate on the hint.) $\endgroup$ – Calvin Lin Nov 17 '20 at 5:24
  • $\begingroup$ @CalvinLin thanks i edited $\endgroup$ – Albus Dumbledore Nov 17 '20 at 5:31
  • 1
    $\begingroup$ I like this a lot! Thank you! Is my understanding for part b correct? $\endgroup$ – Future Math person Nov 17 '20 at 6:06
  • 1
    $\begingroup$ @FutureMathperson yes thats right $\endgroup$ – Albus Dumbledore Nov 17 '20 at 6:08

Another way: By AM-GM $$x^4+x^4+x^4+1\ge 4|x|^3$$ thus $$3(x^4-x^3+1)=\underbrace{(3x^4+1-3x^3)}_{\ge 0}+2>0$$

Here we used the fact that $4|x|^3-3x^3\ge 0$


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