Find range of values of k for which f(x) <= k for all real values of x

Question

Find the range of values of k for which f(x) <=k for all real values of x

   f(x) = -2x^2 + 8x + 17 
  -2x^2 + 8x + 17 <= k
   2x^2 - 8x -17 + k >=0

I have done, but unsure of the answer....

  D = b^2 -4ac > 0
    = 64 + 136 -8k >0
    k<200/8

Help please

Answer 1

Rewrite

$$f(x) =-2(x-2)^2+25\le 25$$

Thus, $k\ge 25$.

In your approach, you should set $D\le 0$ because $f(x)$ and $y = k$ do not intersect or are tangential to each other, which leads to the same range for $k$.

Answer 2

Let $f(x) = -2x^2 + 8x + 17$

Then , $$ f(x)' = -4x+8$$ And $$f(x)'' = -4$$

Since $f(x)'' < 0$ , the function attains maximum value at $$ -4x+8 = 0 \implies x=2$$

At $x = 2, $

$$f(2) = (-2.2^2 + 8.2+17) = -8+16+17 = 25$$

So any value of $k\ge 25$ satisfies the relation.