Find all x-intercepts of the function. Express your answer as a list of x-values. f ( x ) = − 45 x 3 + 60 x 2 − 20 x
1 answer:
Answer:
x = 0,
Step-by-step explanation:
Factorise f(x) and equate to zero for x- intercepts
f(x) = - 5x(9x² - 12x + 4) = 0 ( common factor of - 5x )
To factor the quadratic consider the factors of the product of the x² term and the constant term which sum to give the coefficient of the x- term, that is
product = 9 × 4 = 36 , sum = - 12
The factors are - 6 and - 6
Use these factors to split the middle term
9x² - 6x - 6x + 4 = 0 ( factor the first/second and third/fourth terms )
3x(3x - 2) - 2(3x - 2) ( factor out (3x - 2) )
(3x - 2)(3x - 2), hence
f(x) = - 5x(3x - 2)(3x - 2) = 0
equate each factor to zero and solve for x
- 5x = 0 ⇒ x = 0
(3x - 2)² = 0 ⇒ x = of multiplicity 2, indicating turning point
x - intercepts are x = 0,
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Answer:
The dimension of the box is 18.66 in by 2.66 in by 1.17 in.
Step-by-step explanation:
Given that, The cardboard is 21 in. long and 5 in. wide.
Assume,x be the length of the each sides of the square .
Then the length of the box is = (21-2x) in.
The breadth of the box is =(5-2x) in.
The height of the box is = length of side of the square
= x in.
The volume of the box = length × wide × height
=(21-2x)(5-2x)x
=(105-52x+4x²)x
=()
Let
V=
Differentiating with respect to x
Again differentiating with respect to x
To find the dimension of the box, we set V'=0
Applying the quadratic formula , here a=12, b= -104 and c =105
=7.5, 1.17
For x= 7.5 , the wide of the box will be negative.
∴x=1.17 in.
Since at x= 1.17, V''<0 , therefore at x=1.17 in. the volume of the given box will be maximum.
The length of the box is = [21-(1.17×2)]=18.66 in.
The wide of the box is = [5-(1.17×2)]=2.66 in.
The height of the box is = 1.17 in.
The dimension of the box is 18.66 in by 2.66 in by 1.17 in.