We can apply some rules backwards
first
f'(x)=
(x-4)(-2x+4)
-2x^2+12x-16
we know that
![f'(rx^n)=rnx^{n-1}](https://tex.z-dn.net/?f=f%27%28rx%5En%29%3Drnx%5E%7Bn-1%7D)
so therefor maybe
-2x^2=rnx^{n-1}
2=n-1
n=3
rn=-2
3r=-2
r=-2/3
one is
![\frac{-2}{3}x^3](https://tex.z-dn.net/?f=%20%5Cfrac%7B-2%7D%7B3%7Dx%5E3%20)
second part
12x
12x=
![rnx^{n-1}](https://tex.z-dn.net/?f=rnx%5E%7Bn-1%7D)
x^1, 1=n-1
n=2
rn=12
2r=12
r=6
![6x^2](https://tex.z-dn.net/?f=6x%5E2)
is the second bit
last part
-16
-16x^0=
![rnx^{n-1}](https://tex.z-dn.net/?f=rnx%5E%7Bn-1%7D)
0=n-1
n=1
rn=-16
1r=-16
r=-16
-16x^1
so therfor f(x)=
Answer:
6
Step-by-step explanation:
square root of 36 is 6 ^w^
u = 6
Answer:
I think that technically since it is 0, that there can't be a square root of it because you can't ad anything to it, subtract from it, multiply by it, or divide by it.
Step-by-step explanation:
Answer:
whats the question...
Step-by-step explanation:
TWD is cool
Answer:
x=0.8
x=6
Step-by-step explanation:
(5x-4)(x-6)=0
1)5x-4=0
2)x-6=0
1)5x=4
x=4/5 or 0.8
2)x=6