Respuesta :

apologiabiology
we can apply some rules backwards

first
f'(x)=
(x-4)(-2x+4)
-2x^2+12x-16

we know that
[tex]f'(rx^n)=rnx^{n-1}[/tex]
so therefor maybe
-2x^2=rnx^{n-1}
2=n-1
n=3
rn=-2
3r=-2
r=-2/3

one is
[tex] \frac{-2}{3}x^3 [/tex]
second part

12x
12x=[tex]rnx^{n-1}[/tex]
x^1, 1=n-1
n=2
rn=12
2r=12
r=6
[tex]6x^2[/tex] is the second bit

last part
-16
-16x^0=[tex]rnx^{n-1}[/tex]
0=n-1
n=1

rn=-16
1r=-16
r=-16

-16x^1

so therfor f(x)=[tex] \frac{-2}{3}x^3+6x^2-16x [/tex]

Otras preguntas