Find the max and min value of the quadratic expression 3x^2+2x+11
Answers
Given : quadratic expression 3x^2+2x+11
To find : max and min value of the quadratic expression
Solution:
Z = 3x²+2x+11
dZ/dx = 6x + 2
dZ/dx = 0
=> 6x + 2 = 0
=> x = - 2/6
=> x = - 1/3
d²Z/dx² = 6 > 0
=> value is minimum at x = - 1/3
Minimum value = 3 (-1/3)² + 2(-1/3) + 11
= 1/3 - 2/3 + 11
= 32/3
Minimum value = 32/3
There is no maximum value it goes to infinity
Another method
3x² + 2x + 11
= 3 (x² + 2x/3 + 11/3)
= 3 (( x + 1/3)² - 1/9 + 11/3)
= 3 (( x + 1/3)² + 32/9)
= 3(x + 1/3)² + 32/3
(x + 1/3)² can not be - ve
Hence minimum value = 32/3
Maximum value = ∞
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Step-by-step explanation:
By this direct formula we can direct calculate the maximum or minimum value...
1) If a>0 then minimum
2) If a<0 then maximum
