Answer to Question #289246 in Algebra for bookaddict

Question

f(x)=x^2+x-4 for a<=x<=a+3. if the range of the function f is -2<=f(x)<=16, find the possible values of a

Answer 1

≤ ≥

The range of the  function f is -2≤f(x)≤16. Let's solve the system of inequalities

{x²+x-4 ≥-2

x²+x-4≤16

{x²+x-2 ≥0

x²+x-20 ≤0

For inequality x²+x-2 ≥0 equation roots (by Viette's theorem) x=2 and x=-1

We get x≤-2 and x≥1

For inequality x²+x-20 ≤0 equation roots (by Viette's theorem) x=5 and x=-4

We get -5≤x≤4

Let's solve the system:

{x≤-2 and x≥1

-5≤x≤4

Using the interval method

-5≤x≤-2 and 1≤x≤4

For solution -5≤x≤-2 (according to the task a≤x≤a+3), this implies a=-5

For solution 1≤x≤4 (according to the task a≤x≤a+3), this implies a=1

Answer: a=-5; a=1