00:01
For this problem, we are asked first to show that f of x equals x cubed plus bx squared plus c is an increasing function of b squared is less than 3c.
00:10
So we'll differentiate our expression here.
00:12
F prime of x will be equal to 3x squared plus 2bx plus c.
00:19
If we want to show that this is greater than zero, then what that means is we'll first have to find our, essentially we'll first have to find the zeros of the.
00:31
This expression.
00:32
So we can solve this as a quadratic.
00:35
Particularly, we have that f prime of x equals zero.
00:40
If x is equal to, so it's first negative b, quote unquote, so that would be negative 2b plus or minus, or excuse me, negative 2b over 2a, so that's over 6, or we can reduce this to just negative b over 3.
00:59
We have negative b over 3, plus or minus 1 over 2a, so 1 over 6 times the square root of b squared, which turns into 4 times our actual b squared, minus 4a, c.
01:12
So that's going to be minus 4 times 3, so minus 12c.
01:17
We can factor out the common factor of 4 between these two terms in that square root.
01:23
So we have that this will be when x equals negative b over 3, plus or minus 1 .1 ,000...