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chubhunter [2.5K]
3 years ago
11

Please help me on my equation

Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
5 0

Step-by-step explanation:

g(x) has clearly in its core the same function curve.

but it is

1. upside-down

2. moved up from the x-axis by 1 unit

3. moved to the right of the y-axis by 3 units

so, how do we express these 3 attributes in the functional definition ?

1. easily : by flipping the sign. g(x) = -x² is the same function type just upside-down (mirrored into the negative y space).

g(x) = -x²

2. also very easy : a function is moved up or down on the coordinate grid by adding (or subtracting) a constant.

we need to move our function up by 1 unit : we add 1.

that makes currently g(x) = -x² + 1

3. this is the trickiest part. to move a function left or right we need to make the function "think" that the input value x is not x, but it is (x ± constant).

let's ignore 1. and 2. for the moment and just focus moving the original function 3 units to the right.

that tells us that the functional result value of x in the shifted function must be the same as the functional result value in the original function for an input value that is 3 units "earlier" on the x-axis.

that would mean g(0) = f(-3), g(1) = f(-2), g(2) = f(-1), g(3) = f(0), ...

so, we see, g(x) = f(x-3) = (x-3)²

now, we combine again 1., 2. and 3., and we get

g(x) = -f(x-3) + 1 = -(x-3)² + 1

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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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