Hello, change_for_better!
I agree with your intervals . . .
The graph of the function: .$\displaystyle f(x) \;=\;x^4+x^3+x$ is concave up on the interval:
. . $\displaystyle (a)\;(\infty, \infty) \qquad(b)\text{ None} \qquad (c)\;(1,\infty) \qquad (d)\;(\infty,0) \;\cup\; (1,\infty) \qquad(e)\;(0,\infty)$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I solved and my answer is: .$\displaystyle \left(\infty,\tfrac{1}{2}\right) \cup (0,\infty)$
and I'm sure from my answer but my answer is not included in the choices.
and when I use the graphics calculator the graph is concave up on all interval
. . Um ... not quite!
So what is the correct answer? Your intervals are correct!
$\displaystyle \begin{array}{ccc}f(x) &=&x^4+x^3+x \\ f'(x) &=& 4x^3+3x^2+1 \\ f''(x) &=& 12x^2+6x \end{array}$
$\displaystyle f''(x) > 0 \quad\Rightarrow\quad x < \text{}\tfrac{1}{2}\:\text{ or }\:x > 0$
If you zoom in on your graph, there is a concavedown portion. Code:

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*  *
         o        
* o 
* o 
* * 
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And it's right where you predicted! . . . $\displaystyle \left(\text{}\tfrac{1}{2},\:0\right)$