Answer
A: $f′(x)=\frac{3x^{4}-1}{(x^{4}+x+1)^{2}\sqrt{\frac{\left(x^{4}-x+1\right)}{\left(x^{4}+x+1\right)}}}$
B: $f(x)$ has a horizontal tangent at $x=-0.76$ and $x=0.76$.
C: The graphs of $f(x)$ and $f′(x)$ are consistent with the work done on part B
Work Step by Step
A:
$f(x)=\sqrt{\frac{\left(x^{4}-x+1\right)}{\left(x^{4}+x+1\right)}}$
When we plug in $f(x)$ into a CAS (https://www.derivative-calculator.net/ for this question), we get:
$f′(x)=\frac{3x^{4}-1}{(x^{4}+x+1)^{2}\sqrt{\frac{\left(x^{4}-x+1\right)}{\left(x^{4}+x+1\right)}}}$
B:
To find the horizontal tangents of $f(x)$, we look for when $f(x)$ has local extremas, which also occur when $f`(x)=0$, which is $x=-0.76$ and $x=0.76$.
C:
Using https://www.desmos.com/calculator this time as a CAS to graph, we find that $f(x)$ has a horizontal tangent at $x=-0.76$ and $x=0.76$,. We also see $f′(x)=0$ at $x=-0.76$ and $x=0.76$, which is what proves $f(x)$ has a horizontal tangent.
The graphs of $f(x)$ and $f′(x)$ are consistent with the work done on part B
