Answer:
The intervals of increase are [tex]\left(-4, 0\right) \cup \left(4, \infty\right)[/tex]
Step-by-step explanation:
When a function is increasing, its derivative is positive. So if we want to find the intervals where a function increases, we differentiate it and find the intervals where its derivative is positive.
Let's find the intervals where [tex]f(x) = x^4 -32x^2 + 5[/tex] is increasing.
First, we differentiate f(x)
[tex]\frac{d}{dx} f = \frac{d}{dx} (x^4 -32x^2 + 5)\\\frac{d}{dx} f = \frac{d}{dx}\left(x^4\right)-\frac{d}{dx}\left(32x^2\right)+\frac{d}{dx}\left(5\right)\\\frac{d}{dx} f =4x^3-64x[/tex]
Now we want to find the intervals where [tex]\frac{d}{dx} f[/tex] is positive This is done using critical points, which are the points where [tex]\frac{d}{dx} f[/tex] is either 0 or undefined.
[tex]4x^3-64x=0\\4x\left(x^2-16\right)=0\\4x\left(x+4\right)\left(x-4\right)=0\\\\\mathrm{The\:solutions\:are}\\x=0,\:x=-4,\:x=4[/tex]
These points divide the number line into four intervals
[tex](-\infty,-4);(-4,0);(0,4);(4,\infty)[/tex]
Let's evaluate [tex]\frac{d}{dx} f[/tex] at each interval to see if it's positive on that interval.
[tex]\left\begin{array}{cccc}Interval&x-value&\frac{d}{dx}f& Verdict\\x<-4&-5&-180&Decreasing\\-4<x<0&-1&60&Increasing\\0<x<4&1&-60&Decreasing\\x>4&5&180&Increasing\end{array}\right[/tex]
The intervals of increase are [tex]\left(-4, 0\right) \cup \left(4, \infty\right)[/tex]
