Find the missing values in the triangle below
1 answer:
You might be interested in
I let the computer do the graphing.
f(x) = x(x + 3)²(1 - x) / 3
We see obvious zeros at x=0, x=-3 and x=1.
f(x) = x(x + 3)²(1 - x) / 3
f(x) = x(x^2 + 6x + 9)(1 - x)/3
f(x) = x(x^2 + 6x + 9 - x^3 - 6x^2 - 9x)/3
f(x) = x(- x^3 - 5x^2 -3x +9)/3
f(x) = (-x^4 - 5x^3 -3x^2 +9x)/3
f'(x) = (-4x^3 - 15x^2 -6x + 9)/3
We know there's f'(-3)=0 so x+3 must be a factor here,
f'(x) = (x + 3) (-4x^2 - 3x + 3)
We get stationary points when x = -3 or
x = (-1/8)( 3 ± √(9 - 4(-4)(3))) = (-3 ± √57)/8 about -1.3 and 0.6 which are a local minimum and maximum respectively.
f''(x) = (-12x^2 - 30x -6)/3 = -4x^2 - 10x - 2
f''(-3) = -4(9)+ 30 - 2 = -8
Concave Down, Negative. (-3,0) is a local maximum.
f''((-3 - √57)/8) = -4(-1.3)^2 - 10(-1.3) - 2 = 4.24
Concave up, positive, a CUP
(-3 - √57)/8, f((-3 - √57)/8) ) is a minimum.
f''((-3 + √57)/8) = -4(.6)^2 - 10(.6) - 2 = -9.44
CDN, ( ((-3 + √57)/8), f((-3 + √57)/8) ) is a maximum.
I didn't evaluate f((-3 ±√57)/8) which are around -3 and 1 but I leave that calculation to you.
Answer:
f=11
Mode=4
Median=4
Explanation:
We are given that
a.Mean of the exam score,=3.5
Score(x) frequency C.F
1 1 1
2 3 4
3 f 15(4+f=4+11)
4 13 28
5 4 32
Using the formula
b.Mode:The number which is repeat most times .
4 repeat most times
Hence, mode of all exam scores=4
N=32
N is even
Median=