Can we say that $x^4$ is concave upwards for all $x\in\mathbb{R}$ As we know the second derivative of this function is zero at $x = 0$, but $x = 0$ is not a point of inflection as a point of inflection is a point about which concavity changes but here concavity is same on left and right of $x = 0$ which is upwards, so $x = 0$ in not point of inflection, and as function is defined at $x = 0$ so can we say conclude by this that function is concave upwards for all x.
 A: Given function :$f(x) = x^4 \implies f’(x) = 4x^3 $ It can be seen that derivative of the function is monotonically non-decreasing then on following properties of concave up or convex function, it can be concluded that it is all concave up function.
A: The second derivative test says that a twice-differentiable function $f$ is concave upward on an interval if and only if $f''(x) \geq 0$ for all $x$ in that interval. The second derivative of $x^4$ is $12x^2$, which is clearly non-negative on $\mathbb{R}$. Therefore the function is concave upward on $\mathbb{R}$.
A: The graph of a function is concave up wherever $f''(x) > 0$, which is the case here for all $x \neq 0$. However, when we apply the formal definition of an increasing  function to $f'(x)$, it follows that $f'(x)$ is increasing over $R$; and so, $f(x) = x^{4}$ is concave up over $R$.  
A: Depends on your definition of convexity, I think. The definition I prefer is to say that $f$ is concave up on $[a,b]$ when for every $x_1,x_2,x_3\in[a,b]$ with $a\le x_1<x_2<x_3\le b$, the slope from $\bigl(x_1,f(x(1)\bigr)$ to $\bigl(x_2,f(x_2)\bigr)$ is less than the slope from $\bigl(x_1,f(x(1)\bigr)$ to $\bigl(x_3,f(x_3)\bigr)$. Write out the inequality and simplify and get the condition
$$
x_1[f(x_2) - f(x_3)]\, +\, x_2[f(x_3) - f(x_1))]\, + \,x_3[f(x_1)-f(x_2)]\>>\>0\,.
$$
With this definition, $f(x)=x^{2n}$ is concave upwards.
Note that the above definition does not mention derivatives, so it’s perfectly good for non-differentiable functions, and does not say what convexity at a point might be.
