Determine the maximum value of y=3x\(^2\) + 5x - 3
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Correct Answer: Option D
Explanation:
y=3x\(^2\) + 5x - 3
dy/dx = 6x + 5
as dy/dx = 0
6x + 5 = 0
x = \(\frac{-5}{6}\)
Maximum value: 3 \( ^2{\frac{-5}{6}}\) + 5 \(\frac{-5}{6}\) - 3
3 \(\frac{75}{36}\) - \(\frac{25}{6}\) - 3
Using the L.C.M. 36
= \(\frac{25 - 50 - 36}{36}\)
= \(\frac{-61}{36}\)
No correct option
y=3x\(^2\) + 5x - 3
dy/dx = 6x + 5
as dy/dx = 0
6x + 5 = 0
x = \(\frac{-5}{6}\)
Maximum value: 3 \( ^2{\frac{-5}{6}}\) + 5 \(\frac{-5}{6}\) - 3
3 \(\frac{75}{36}\) - \(\frac{25}{6}\) - 3
Using the L.C.M. 36
= \(\frac{25 - 50 - 36}{36}\)
= \(\frac{-61}{36}\)
No correct option