I've been asked to find the area under the graph of the function $x^4 -2x^3 +2$ between $x=-1$ and $x=2$, I can easily do this but now I would like to confirm that the function is nonnegative on this interval.
Plotting the function is easily seen to be positive everywhere, but I always like to do this algebraically, so best I can think of is checking if the polynomial has any roots, as far as I know all the roots are found using the coefficient of $x^4$ which is $1$ and the independen term which is $2$ now the possible roots are $1,-1,2,-2$ none of this are roots so the poynomial has no real roots. Now since $f(0)=2$ the polinomial does not change sign so it is always positive.
Is this sufficient or is there a better way prove this?
$\endgroup$ 11 Answer
$\begingroup$Examine $f'(x) = 4x^3 - 6x^2 = 2x^2(2x-3)$
From this you can see that $f$ has critical points at $x=0$ and at $x = \dfrac 32$. We can see that $x=0$ is neither a minimum or a maximum, because $f'$ doesn't change its sign from the left to the right of $0$. You can tell $x = \dfrac 32$ is a minimum because this is an even degree polynomial with a positive coefficient, so since there is only one turning point, it must be a minimum. Alternatively, you can use the second (or firs) derivative test to confirm that this is a minimum.
Plugging in $\dfrac 32$ into $f$, we see that $f \left( \dfrac 32 \right)$ is positive, and since this a minimum, the function is positive everywhere.
P.S. Like I said in the comment, ust because $\pm1$ and $\pm 2$ aren't zeroes doesn't mean the polynomial doesn't have real roots, it just means the polynomial doesn't have rational roots.
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