dy=0

if y=(x^2-ax+b)(x+c)

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For the beginning, let's differentiate the given function.

dy=(x^2-ax+b)(x+c)dx

Since the function is a product, we'll apply the product rule, when differentiating a product.

(f*g)' = f'*g + f*g'

We'll put f = x^2-ax+b and g = x+c

We'll differentiate, to the right side, with respect to x:

[ (x^2-ax+b)(x+c) ]' = (x^2-ax+b)' * (x+c) + (x^2-ax+b) * (x+c)'

[ (x^2-ax+b)(x+c) ]' = (2x-a) * (x+c) + (x^2-ax+b) * (1)

We'll remove the brackets:

[ (x^2-ax+b)(x+c) ]' = 2x^2 + 2xc - ax - ac + x^2-ax+b

Now, we'll put dy = 0

We'll substitute the expression for dy:

2x^2 + 2xc - ax - ac + x^2-ax+b = 0

We'll combine like terms:

3x^2 + x(2c - 2a) + b - ac = 0

If we'll plug in values for the a,b,c, we'll calculate the values of x for dy = 0.

Since the expression is a quadratic equation, we'll have the following cases:

For the quadratic to have 2 distinct roots, we'll have to impose the constraint: delta > 0

delta = (2c - 2a)^2 - 12( b - ac)

delta = 4c^2 - 8ac + 4a^2 - 12b + 12ac

delta = 4a^2 + 4c^2 + 4ac - 12b > 0

We'll divide by 4:

a^2 + c^2 + ac - 3b > 0

From this expression, we'll conclude that: a,c>b

For the quadratic to have 2 equal roots, we'll have to impose the constraint: delta = 0.

a^2 + c^2 + ac - 3b= 0

For the quadratic not to have any roots, we'll have to impose the constraint: delta < 0.

a^2 + c^2 + ac - 3b<0

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