Science:Math Exam Resources/Courses/MATH104/December 2014/Question 05 (d)/Solution 1

Recall that when $f''>0$, $f$ is concave up; likewise, when $f''<0$, $f$ is concave down. So what we need to do is find the intervals on which $f''$ exists and is non-zero. From above, we know that $f''(x)={\frac {e^{x}(x^{2}-4x+6)}{x^{4}}}$, and indeed, $f''(x)=0$ if and only if $e^{x}=0$ or $x^{2}-4x+6=0$. Since $e^{x}>0$ for all $x$, we have that $x^{2}-4x+6=0$. Using the quadratic formula, we find that this quadratic equation has no real roots. Therefore, there is no $x$ where $f''(x)=0$. Now, we make a table as before for $f'$, noting that $f''$ does not exist at $x=0$. This gives us two intervals, namely $(-\infty ,0)$ and $(0,\infty )$ where $f''$ is nonzero and defined:

	$(-\infty ,0)$	$(0,\infty )$
$e^{x}$	+	+
$x^{2}-4x+6$	+	+
$x^{4}$	+	+
$f''$	+	+
$f$	Concave Up	Concave Up

So on both $(-\infty ,0)$ and $(0,\infty )$, $f$ is concave up.