+30 For a real number x, let
\(f(x) = \left\{ \begin{array}{cl} x^2 & \text{for x ≤ -2} \\ |2x + 1| & \text{for -2 < x < 3} \\ 3x+8 & \text{for x ≥ 3} \\ \end{array} \right.\)
Find all x for which f(x)=5.
\(\)
+128475 x^2 = 5
x = -sqrt 5 or sqrt 5
Since -sqrt 5 ≤ -2
Then x = -sqrt 5 is one solution
And
2 x + 1 = 5 or 2x + 1 = -5
2x = 4 2x = -6
x = 2 x = -3
Since x = 2 is on the interval -2< x < 3
Then x = 2 is another solution
And
3x + 8 = 5
3x = -3
x = -1
Not a solution since x = -1 not on the defined interval for this function
So
f(-sqrt 5) = 5
and
f(2) = 5
See the graph here : https://www.desmos.com/calculator/fcqghft5mb
