How to evaluate a function with a quadratic argument or higher? How to proceed?example if I have a function of the form f ( 2 x + 5 ) = 4 x − 3and I am asked to evaluatef(8)I make t = 2 x + 5from where x = ( t − 5 ) / 2then f ( t ) = 4 ( ( t − 5 ) / 2 ) − 3I take t = 8and that's it.**Doubtsbut how do I do it here, if I do the above I get a quadratic equation or a cubic onea) f ( x 2 + 1 / x 2 ) = x + 1 / xfind f ( 4 ) = ? x &gt; 0 b) f ( x 3 + 1 / x 3 ) = x + 1 / xfind f ( 4 ) = ? x &gt; 0

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How to evaluate a function with a quadratic argument or higher? How to proceed?example if I have a function of the form  f  (  2  x  +  5  )  =  4  x  −  3and I am asked to evaluatef(8)I make  t  =  2  x  +  5from where  x  =  (  t  −  5  )      /    2then  f  (  t  )  =  4  (  (  t  −  5  )      /    2  )  −  3I take  t  =  8and that&#039;s it.**Doubtsbut how do I do it here, if I do the above I get a quadratic equation or a cubic onea)   f  (      x    2    +  1      /        x    2    )  =  x  +  1      /    xfind  f  (  4  )  =  ?  x  &amp;gt;  0 b)  f  (      x    3    +  1      /        x    3    )  =  x  +  1      /    xfind  f  (  4  )  =  ?  x  &amp;gt;  0

Falpo359 · Accepted Answer

Step 1Sometimes it helps to see things a different way. Instead of solving for x, I would have looked at the equation as  f  (  2  x  +  5  )  =  4  x  −  3  =  4  x  +  10  −  10  −  3  =  2  (  2  x  +  5  )  −  13which means   f  (  x  )  =  2  x  −  13. Taking the second equation we have  f      (          x      2        +          1              x        2              )    =  x  +      1    x    =                    (        x        +                  1          x                )            2        =            x      2        +    2    +          1              x        2              =            (              x        2            +              1                  x          2                    )        +    2  which means   f  (  x  )  =      x    +    2   if   x  &amp;gt;  0 (you will need to clarify whether this is the case).

sbrigynt7b · Accepted Answer

Step 1Just take that as:Find   x  +            1      x       given       x    2    +            1              x        2              =  4.And this is easily solvable by algebraic methods.Step 2Find   x  +            1      x       given       x    3    +            1              x        3              =  4.Let   x  +            1      x        =  k. Then,We know,  (  a  +  b      )    3    =      a    3    +      b    3    +  3  a  b  (  a  +  b  )Step 2Putting   a  =  x and   b  =            1      x      , we get      k    3    =      x    3    +            1              x        3              +  3  ×  x  ×            1      x        ×  k      k    3    −  3  k  −  4  =  0Solving gives  k  =                                                        2              −                              3                                      3                    +                                    2              +                              3                                      3

How to evaluate a function with a quadratic argument or higher? How to proceed? example if I have a function of the form f(2x+5)=4x-3 and I am asked to evaluate f(8)

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