Question:
For which values of 'a' will the function $f(x)=x^4+ax^3+\frac3 2 x^2+1$ will be concave upward along the entire real line.
My Approach:
I know that the concavity (concave upwards or concave downwards) is determined by the sign of the second derivative of the function.
So I computed the second derivate as follows:
$f'(x)=4x^3+3ax^2+3x$
$f''(x)=12x^2+6ax+3$
For the given function $f(x)$ to be concave upwards for all real values of $x$, the second derivative must always be positive.
We see that for $f''(x)$ the coefficient of $x^2$ is $12$ which is positive. So in order for the second derivative to be positive for all real values of $x$, I solved the inequality, Discriminant of the second derivative to negative and obtained the following:
$36a^2-36*4<0$
$\implies a^2-4<0$
$\implies a^2<4$
$\implies a \in (-2,+2)$ or in other words $a$ can take all real values between $-2$ and $+2$, excluding both $-2$ and $+2$.
But the answer in my book states $a \in [-2,+2]$ both $-2$ and $+2$ included.
I counter checked the answer using graphing calculator (Desmos) as follows:
For $a=-2$,
For $a=+2$,
Clearly, for both the values of $a$ the graph of the function(black curve) is concave upwards.
So my doubt is why should we consider the case the discriminant of the second derivative to be "less than and equal to zero"(the correct answer in book as well as one verified using graphs) rather than the case it is "less than zero"(as I did).
Thank you in advance.