IJ=x-4
JK=5x+12
IK=9x-19
IJ+JK=IK
(x-4)+(5x+12)=(9x-19)
x-4+5x+12=9x-19
6x+8=9x-19
6x+8-6x+19=9x-19-6x+19
27=3x
3x=27
3x/3=27/3
x=9
IJ=x-4=9-4→IJ=5
JK=5x+12=5(9)+12=45+12→JK=57
IK=9x-19=9(9)-19=81-19→IK=62
Answer: Option B. IJ=5, JK=57, IK=62
Answer:
- local minimum at x = -3-2√14
- local maximum at x = -3+2√14
Step-by-step explanation:
The first derivative simplifies to ...
d/dx( ) = -(x^2 +6x -47)/(x^2 -9x +20)^2
This has zeros that can be found by the usual methods of solving quadratics:
x = -3 ±2√14
The positive value of x corresponds to what I might call an "apex." (See the first attachment.) The negative value is where the function turns around an approaches the horizontal asymptote from below. (See the second attachment.)
Plug in 5 for n so it would be 5×5×5×5=
C.) 625