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Mariana [72]
3 years ago
9

Find f'(x) f(x)=(1-2x^2)^3

Mathematics
1 answer:
Hunter-Best [27]3 years ago
3 0

Answer:  f'(x) = 12x  + 48x³ - 48x⁵

<u>Step-by-step explanation:</u>

f(x) = (1 - 2x²)³

     = (1 - 2x²)(1 - 4x² + 4x⁴)

         1 - 4x² + 4x⁴

        <u>    -2x² + 8x⁴ - 8x⁶</u>

     =  1 - 6x² + 12x⁴ - 8x⁶

f'(x) = 0 -2(6)x²⁻¹ + 4(12)x⁴⁻¹ - 6(8)x⁶⁻¹

      =     12x  + 48x³ - 48x⁵

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Find all the zeros for each function P(x)=x^4-4x^3-x^2+20x-20
ivanzaharov [21]

Answer:

The zeros of the given polynomial function are

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Step-by-step explanation:

Given polynomial is P(x)=x^4-4x^3-x^2+20x-20

To find the zeros equate the given polynomial to zero

ie., P(x)=0

P(x)=x^4-4x^3-x^2+20x-20=0

By using synthetic division we can solve the polynomial:

2_|   1     -4     -1      20      -20

       0      2     -4     -10       20

   _____________________

       1     -2      -5      10      |_0

Therefore x-2=0

x=2 is a zero of P(x)

Now we can write the cubic equation as below:

x^3-2x^2-5x+10=0

Again using synthetic division

2_|   1     -2     -5     10      

       0      2      0    -10    

    ______________

       1      0      -5     |_0

Therefore x-2=0

x=2 is also a zero of P(x).

Now we have x^2+0x-5=0

x^2-5=0

x^2=5

x^=\pm\sqrt{5} is a zero of P(x)

Therefore the zeros are 2,2,\pm\sqrt{5}

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0.25 km is the answer.

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