Consider the function
![f(x)=x^{1/3}](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E%7B1%2F3%7D)
, which has derivative
![f'(x)=\dfrac13x^{-2/3}](https://tex.z-dn.net/?f=f%27%28x%29%3D%5Cdfrac13x%5E%7B-2%2F3%7D)
.
The linear approximation of
![f(x)](https://tex.z-dn.net/?f=f%28x%29)
for some value
![x](https://tex.z-dn.net/?f=x)
within a neighborhood of
![x=c](https://tex.z-dn.net/?f=x%3Dc)
is given by
![f(x)\approx f'(c)(x-c)+f(c)](https://tex.z-dn.net/?f=f%28x%29%5Capprox%20f%27%28c%29%28x-c%29%2Bf%28c%29)
Let
![c=64](https://tex.z-dn.net/?f=c%3D64)
. Then
![(63.97)^{1/3}](https://tex.z-dn.net/?f=%2863.97%29%5E%7B1%2F3%7D)
can be estimated to be
![f(63.97)\approxf'(64)(63.97-64)+f(64)](https://tex.z-dn.net/?f=f%2863.97%29%5Capproxf%27%2864%29%2863.97-64%29%2Bf%2864%29)
![\sqrt[3]{63.97}\approx4-\dfrac{0.03}{48}=3.999375](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B63.97%7D%5Capprox4-%5Cdfrac%7B0.03%7D%7B48%7D%3D3.999375)
Since
![f'(x)>0](https://tex.z-dn.net/?f=f%27%28x%29%3E0)
for
![x>0](https://tex.z-dn.net/?f=x%3E0)
, it follows that
![f(x)](https://tex.z-dn.net/?f=f%28x%29)
must be strictly increasing over that part of its domain, which means the linear approximation lies strictly above the function
![f(x)](https://tex.z-dn.net/?f=f%28x%29)
. This means the estimated value is an overestimation.
Indeed, the actual value is closer to the number 3.999374902...
Answer: Increase in number or size, at a constantly growing rate. It is one possible result of a reinforcing feedback loop that makes a population or system grow (escalate) by increasingly higher amounts.
First solve the return in each
price per unit
Return = number units sold x price
per unit
Return1 = 5000 units x Php 900
Return1 = php 4,500,000
Next is solve the other price per
unit
Return2 = number units sold x
price per unit
Return2 = (5000 + 1500 units0 x (
php 900 – 100)
Return2 = php 5,200,000
So the php 800 per unit will have
greater return
Answer:
wow why is this so hard
Step-by-step explanation:
i am in 8th grade to be honest