Answer and Explanation:
A function is said to be increasing, if the derivative of function is f’(x) > 0 on each point. A function is said to be decreasing if f”(x) < 0.
Let y = v (z) be differentiable on the interval (a, b). If two points z1 and z2 belongs to the interval (a, b) such that z1 < z2, then v (z1) ≤ v (z2), the function is increasing in this interval.
Similarly, the function y = v(z) is said to be decreasing, when it is differentiable on the interval (a , b).
Two points z1 and z2 Є (a, b) such that z1 > z2, then v (z1) ≥ v(z2). The function is decreasing on this interval.
The function y = v (z)
The derivative of function Y’ = v’(z) is positive, then the function is increasing.
The function y = v (z)
The derivative of function y’ is negative, then the function is decreasing.
The increasing and the decreasing intervals of the function V(z) are
[tex]\mathrm{Increasing}:-\infty \: < z < 0,\:\\\mathrm{Decreasing}:0 < z < \frac{16}{7},\:\\\mathrm{Increasing}:\frac{16}{7} < z < 4,\:\\\mathrm{Increasing}:4 < z < \infty \:[/tex]
The function is given as:
[tex]V(z)=z^4(2z-8)^3[/tex]
To determine the increasing and the decreasing interval, we plot the graph of V(z)
From the graph, we have the following highlights
[tex]\mathrm{Increasing}:-\infty \: < z < 0,\:\\\mathrm{Decreasing}:0 < z < \frac{16}{7},\:\\\mathrm{Increasing}:\frac{16}{7} < z < 4,\:\\\mathrm{Increasing}:4 < z < \infty \:[/tex]
The above represent the increasing and the decreasing intervals of the function V(z)
Read more about increasing and decreasing intervals at:
https://brainly.com/question/1503051
