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lorasvet [3.4K]
3 years ago
9

If f'(x) is (x-4)(4-2x) , what is f(x)?

Mathematics
1 answer:
butalik [34]3 years ago
3 0
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}
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
second part

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

last part
-16
-16x^0=rnx^{n-1}
0=n-1
n=1

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

-16x^1

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