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sdas [7]
2 years ago
7

Gradient from 2 - 1/2y = 0​

Mathematics
1 answer:
kirill115 [55]2 years ago
4 0

Given :

  1. 2 - 1/2y = 0

To find :-

  • The gradient form .

Solution :-

<u>The </u><u>gradient</u><u> form</u><u>/</u><u> </u><u>slope</u><u> intercept</u><u> form</u><u> </u>

  • y = mx + c ,

<u>On </u><u>converting</u><u> </u><u>,</u>

  • 2 - 1/2y = 0
  • 1/2y = 2
  • y = 2*2
  • y = 4
  • y = 0x + 4
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2 years ago
F(x)=2x an$ g(x)=2x+3,what is the value of f(g(-8))
Fudgin [204]

First plug in g(x) into f(x)

F((g(x))=2(2x+3)

And now you plug in -8

F(g(-8))=2(-16+3)

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F(g(-8))=-26

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2 years ago
Please see attachment
Dafna11 [192]

Answer:

a) The value of absolute minimum value = - 0.3536  

b) which is attained at   x = \frac{1}{\sqrt{2} }  

Step-by-step explanation:

<u>Step(i)</u>:-

Given function

                       f(x) = \frac{-x}{2x^{2} +1}     ...(i)

Differentiating equation (i) with respective to 'x'

                     f^{l} = \frac{2x^{2} +1(-1) - (-x) (4x)}{(2x^{2}+1)^{2}  }   ...(ii)

                    f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2}  }

Equating Zero

                   f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2}  } = 0

                 \frac{2x^{2}-1}{(2x^{2}+1)^{2}  } = 0

                2 x^{2}-1 = 0

               2 x^{2} = 1

             x^{2}  = \frac{1}{2}

             x = \frac{-1}{\sqrt{2} }  , x = \frac{1}{\sqrt{2} }

<u><em>Step(ii):</em></u>-

Again Differentiating equation (ii) with respective to 'x'

f^{ll}(x) = \frac{(2x^{2} +1)^{2} (4x) - 2(2x^{2} +1) (4x)(2x^{2}-1) }{(2x^{2}+1)^{4}  }

put

      x = \frac{1}{\sqrt{2} }

f^{ll} (x) > 0

The absolute minimum value at   x = \frac{1}{\sqrt{2} }

<u><em>Step(iii):</em></u>-

The value of absolute minimum value

                         f(x) = \frac{-x}{2x^{2} +1}

                       f(\frac{1}{\sqrt{2} } ) = \frac{-\frac{1}{\sqrt{2} } }{2(\frac{1}{\sqrt{2} } )^{2} +1}

         on calculation we get

The value of absolute minimum value = - 0.3536      

<u><em>Final answer</em></u>:-

a) The value of absolute minimum value = - 0.3536  

b) which is attained at   x = \frac{1}{\sqrt{2} }    

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