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secondary 4 | A Maths
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secondary 4 chevron_right A Maths chevron_right Singapore

y = 2x^3 -9x^2+5

Date Posted: 2 years ago
Views: 570
Eric Nicholas K
Eric Nicholas K
2 years ago
y = 2x3 - 9x2 + 5

The gradient function of the curve is given by

dy/dx = 6x2 - 18x

The gradient itself is an expression in x; the minimum value of the gradient, or the minimum value of this expression 6x2 - 18x, is found by differentiating this expression one more times and setting the new expression is zero

[For a minimum value of something, its derivative is equal to zero]

d2y/dx2 = 12x - 18

So,
12x - 18 = 0
12x = 18
x = 3/2

Minimum gradient
= 6 (3/2)^2 - 18 (3/2)
= 27/2 - 27
= -27/2
= -13 1/2
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2 years ago
i did the same thing but my tuition cher said that im wrong and i have to use first derivative test
Eric Nicholas K
Eric Nicholas K
2 years ago
Technically you still have to do that on the d2y/dx2 of 12x - 18, but it’s too obvious for a function.

Or you could try d3y/dx3 = 12 and this indicates that the particular value of dy/dx must certainly be a minimum.

Alternatively, you could complete the square on dy/dx

dy/dx
= 6x2 - 18x
= 6 (x2 - 3x)
= 6 [x2 - 3x + 1.5^2 - 1.5^2]
= 6 [(x - 1.5)^2 - 2.25]
= 6 (x - 1.5)^2 - 13.5

And you can see that this “quadratic” expression is going to have a clear minimum of -13.5 when x = 1.5.
Eric Nicholas K
Eric Nicholas K
2 years ago
Note that the “first derivative test” here is testing for values of d2y/dx2 around the stationary point, not the values of dy/dx around the stationary point, because the derivative of the gradient function is called d2y/dx2.

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Arnold K H Tan
Arnold K H Tan's answer
2155 answers (A Helpful Person)
1st
The easiest is to complete the square for dy/dx to determine the minimum gradient. Alternative is to use the first derivative sign test or to use d²y/dx²