If a is real, then the Range of 2x^ 2 -6x+8 is
Answers
Range of 2x² - 6x + 8 is [7/2, ∞) if x is real
Given:
- x is real
To Find :
- Range of 2x² - 6x + 8
Assume that f(x) = 2x² - 6x + 8
Concept to be used :
To find Minimum/Maximum value derivative of f(x) to be found.
f(x) = 2x² - 6x + 8
Step 1 :
Taking derivative
f'(x) = 4x - 6
Step 2 :
Equate derivative with 0
=> 4x - 6 = 0
Step 3 :
Solve for x
=> x = 6/4 = 3/2
Step 4 :
Find 2nd derivative
f''(x) = 4 > 0
Hence f(x) minimum value at x = 3/2
Step 5 :
Calculate f(x) at x = 3/2
= 2(3/2)² - 6(3/2) + 8
= 4.5 - 9 + 8
= 3.5
= 7/2
Minimum value is 7/2 and no maximum value
Hence Range of 2x² - 6x + 8 is [7/2, ∞)
(There was a typing mistake in the Question , probably a is real should be x is real)
Another method
2x² - 6x + 8
Step 1 :
Taking 2 as common factor
= 2(x² - 3x + 4)
Step 2:
Completing the square
= 2((x - 3/2)² - 9/4 + 4)
= 2(x - 3/2)² + 7/2
As (x - 3/2)² can not be negative Hence minimum value = 7/2
And there is not limit for maximum value.
Hence Range of 2x² - 6x + 8 is [7/2, ∞)