# Showing that the graph of a function lies entirely above the x-axis

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

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

• nice(+1) but how did you come up with that,i am curious – Albus Dumbledore Nov 17 '20 at 5:42
• @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. – Calvin Lin Nov 17 '20 at 5:44
• i see ,thanks for insight am-gm also helps as i did below – Albus Dumbledore Nov 17 '20 at 5:46
• This was my original plan but I didn't know how to go about doing this. Thank you! – 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

• +1 (Though you might want to elaborate on the hint.) – Calvin Lin Nov 17 '20 at 5:24
• @CalvinLin thanks i edited – Albus Dumbledore Nov 17 '20 at 5:31
• I like this a lot! Thank you! Is my understanding for part b correct? – Future Math person Nov 17 '20 at 6:06
• @FutureMathperson yes thats right – 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 $$4|x|^3-3x^3\ge 0$$