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