1 Answer | Add Yours
For the beginning, let's differentiate the given function.
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
We’ve answered 317,879 questions. We can answer yours, too.Ask a question