Let f(x) = (1-4x^2)/(x)
For what values of x is f decreasing? For what values of x is the graph of f concave down?

Question

Let f(x) = (1-4x^2)/(x)
For what values of x is f decreasing? For what values of x is the graph of f concave down?

tlar · Answer

@Badsheep

solomonzelman · Answer

1.   $\color{black}{\displaystyle f(x)  }$ increases on $\color{black}{\displaystyle I  }$, iff $\color{black}{\displaystyle f'(x)>0  }$ on $\color{black}{\displaystyle I  }$.
2.   $\color{black}{\displaystyle f(x)  }$ decreases on $\color{black}{\displaystyle I  }$, iff $\color{black}{\displaystyle f'(x)<0  }$ on $\color{black}{\displaystyle I  }$.

solomonzelman · Answer

So, to find $\color{black}{{\rm for~what~values~}f~{\rm is~decreasing}}$, all you need to do find $\color{black}{\displaystyle f'(x)  }$ and to then set $\color{black}{\displaystyle f'(x)<0  }$ and solve for $\color{black}{\displaystyle x  }$.

solomonzelman · Answer

The relation b/w $\color{black}{\displaystyle f  }$ and $\color{black}{\displaystyle f'  }$, is the same relation as that b/w $\color{black}{\displaystyle f'  }$ and $\color{black}{\displaystyle f''}$.

In other words,

1b).   $\color{black}{\displaystyle f'(x)  }$ (or the slope of $\color{black}{\displaystyle f  }$) increases on $\color{black}{\displaystyle I  }$, iff $\color{black}{\displaystyle f''(x)>0  }$ on $\color{black}{\displaystyle I  }$.
2b).   $\color{black}{\displaystyle f'(x)  }$ (or the slope of $\color{black}{\displaystyle f  }$) decreases on $\color{black}{\displaystyle I  }$, iff $\color{black}{\displaystyle f''(x)<0  }$ on $\color{black}{\displaystyle I  }$.

solomonzelman · Answer

What "concave down" means, is that the (not the $f$, but the) slope of $f$ is decreasing. So, in order to find where $f$ is concave down, you need to find $f''(x)$ and then solve $f''(x)<0$.