Find the range of values of k such that f(x) is ontoI am trying to solve the below problem in pre-calculus, I give below the steps I followed , I am stuck as it gets very complicated , need help in solution.A function f : R → R is defined by f ( x ) = k x 2 + 6 x − 8 k + 6 x − 8 x 2 .Find the intervals of values of k such that f is onto. The answer given in the book is 2 ≤ k ≤ 14.I started by trying to find the range of the function and find the values of k such that the range is = R ( that is range = co domain , in this case co domain is R and hence has to prove that range = R)Let y = k x 2 + 6 x − 8 k + 6 x − 8 x 2 Rearranging and grouping gives ( k + 8 y ) x 2 + 6 ( 1 − y ) x − ( 8 + k y ) = 0.This is quadratic in x and for x to be real d e t ( x ) ≥ 0, which requires 36 ( 1 − y ) 2 + 4 ( k + 8 y ) ( 8 + k y ) ≥ 0.Rearranging and grouping this equation gives ( 36 + 32 k ) y 2 + ( 4 k 2 + 184 ) y + ( 36 + 32 k ) ≥ 0.Now I have to find the range of values of k such that the above expression is ≥ 0. Totally stuck here. Please help. Also, is there any other easier way of arriving at the range of k ?

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Find the range of values of k such that f(x) is ontoI am trying to solve the below problem in pre-calculus, I give below the steps I followed , I am stuck as it gets very complicated , need help in solution.A function   f  :      R    →      R   is defined by  f  (  x  )  =            k              x        2            +      6      x      −      8              k      +      6      x      −      8              x        2              .Find the intervals of values of k such that f is onto. The answer given in the book is   2  ≤  k  ≤  14.I started by trying to find the range of the function and find the values of k such that the range is = R ( that is range = co domain , in this case co domain is R and hence has to prove that range = R)Let  y  =            k              x        2            +      6      x      −      8              k      +      6      x      −      8              x        2            Rearranging and grouping gives  (  k  +  8  y  )      x    2    +  6  (  1  −  y  )  x  −  (  8  +  k  y  )  =  0.This is quadratic in x and for x to be real   d  e  t  (  x  )  ≥  0, which requires  36  (  1  −  y      )    2    +  4  (  k  +  8  y  )  (  8  +  k  y  )  ≥  0.Rearranging and grouping this equation gives  (  36  +  32  k  )      y    2    +  (  4      k    2    +  184  )  y  +  (  36  +  32  k  )  ≥  0.Now I have to find the range of values of k such that the above expression is   ≥  0. Totally stuck here. Please help. Also, is there any other easier way of arriving at the range of k ?

retalibry9 · Accepted Answer

Step 1Start with your result  (  36  +  32  k  )      y    2    +  (  4      k    2    +  184  )  y  +  (  36  +  32  k  )  ≥  0If we draw a graph about   z  =  (  36  +  32  k  )      y    2    +  (  4      k    2    +  184  )  y  +  (  36  +  32  k  ) which axes are y and z. We want this be a upper-open parabola and not touch the points below the horizontal axis. so what we want is      {                            36          +          32          k          &amp;gt;          0                                      Δ          =          (          4                      k            2                    +          184                      )            2                    −          4          (          36          +          32          k                      )            2                    ≤          0                        First solve the second inequality, using formula of the different of squares, we have                    (        4                  k          2                +        184        +        72        +        64        k        )        (        4                  k          2                +        184        −        72        −        64        k        )                            ≤        0                            (                  k          2                +        16        k        +        64        )        (                  k          2                −        16        k        +        28        )                            ≤        0                            (        k        +        8                  )          2                (        k        −        2        )        (        k        −        14        )                            ≤        0            Step 2Since   (  k  +  8      )    2   will always stay non-negative, it will have only little influence in the inequality, then we just have to solve  (  k  −  2  )  (  k  −  14  )  ≤  0which give us   2  ≤  k  ≤  14. Now see that for all k in this range,   36  +  32  k will always stay positive. Hence the answer is   2  ≤  k  ≤  14.

Find the range of values of k such that f(x) is onto. A function f:R rightarrow R is defined by f(x)=(kx^2+6x-8)/(k+6x-8x^2)

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