f(x) = x^3(3-x)^4
I get derivative = ( 3x^2 * (3-x)^4 ) + ( 4x^3 * (3-x)^3) )
= 12x^5 * (3-x)^4 * (3-x)^3
I get x = 0 & 3
but what is the third x value to test?
The derivative of the function is
f'(x) = 3x² · ( 3 - x)⁴ + x³ · 4(3 - x)³ · ( - 1 ) = 3x² · ( 3 - x)⁴ - 4x³(3 - x)³ =
= x²(3 - x)³[ 3(3 - x) - 4x ] =
= x²·(3 - x)³·(9 - 3x - 4x) = x²·(3 - x)³·(9 - 7x)
The critical points are the solutions of this equation
f'(x) = 0
x²·(3 - x)³·(9 - 7x) = 0
x² = 0 → x = 0
(3 - x)³ = 0 → x = 3
9 - 7x = 0 → x = 9/7
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Verified answer
The derivative of the function is
f'(x) = 3x² · ( 3 - x)⁴ + x³ · 4(3 - x)³ · ( - 1 ) = 3x² · ( 3 - x)⁴ - 4x³(3 - x)³ =
= x²(3 - x)³[ 3(3 - x) - 4x ] =
= x²·(3 - x)³·(9 - 3x - 4x) = x²·(3 - x)³·(9 - 7x)
The critical points are the solutions of this equation
f'(x) = 0
x²·(3 - x)³·(9 - 7x) = 0
x² = 0 → x = 0
(3 - x)³ = 0 → x = 3
9 - 7x = 0 → x = 9/7