Find all intervals on which $f(x)$ is decreasing. \[ f(x)=\frac{1}{5} x^{5}+4 x^{4}+20 x^{3}-31 \]
Solution
Step 1 ：Find the derivative of the function \(f(x)\), which is \(f'(x)\), and it is given by: \[f'(x)=x^{4}+16 x^{3}+60 x^{2}\]
Step 2 ：Set the derivative equal to zero and solve for x. This gives us: \[x^{4}+16 x^{3}+60 x^{2}=0\]
Step 3 ：Factor out \(x^2\) to get: \[x^{2}(x^{2}+16 x+60)=0\]
Step 4 ：Setting each factor equal to zero gives us \(x=0\) and the solutions to the quadratic equation \(x^{2}+16 x+60=0\). However, the discriminant is negative in this case, which means there are no real solutions to the quadratic equation. Therefore, the only critical point is \(x=0\).
Step 5 ：Test the sign of the derivative \(f'(x)\) in the intervals \((-\infty, 0)\) and \((0, \infty)\). If \(f'(x)<0\) in an interval, then \(f(x)\) is decreasing in that interval.
Step 6 ：For \(x<0\), we have \(f'(x)=x^{4}+16 x^{3}+60 x^{2}>0\) because all terms are positive.
Step 7 ：For \(x>0\), we have \(f'(x)=x^{4}+16 x^{3}+60 x^{2}>0\) because all terms are positive.
Step 8 ：\(\boxed{\text{Therefore, the function } f(x) \text{ is not decreasing on any interval.}}\)
