Obtain all other zeroes of the polynomial x4 – x3 – 15x² + 3x +36, if two of its zeroes are 13 and 3
-35 are 2 + V3.Find the other
Answers
Answer:
Sorry I don't know
Step-by-step explanation:
As I am in lower class
Answer:
Example 1 :
Determine maximum values of the functions
y = 4x - x2 + 3
Solution :
f(x) = y = 4x - x2 + 3
First let us find the first derivative
f'(x) = 4(1) - 2x + 0
f'(x) = 4 - 2x
Let f'(x) = 0
4 - 2x = 0
2 (2 - x) = 0
2 - x = 0
x = 2
Now let us find the second derivative
f''(x) = 0 - 2(1)
f''(x) = -2 < 0 Maximum
To find the maximum value, we have to apply x = 2 in the original function.
f(2) = 4(2) - 22 + 3
f(2) = 8 - 4 + 3
f(2) = 11 - 4
f(2) = 7
Therefore the maximum value is 7 at x = 2. Now let us check this in the graph.
Checking :
y = 4x - x2 + 3
The given function is the equation of parabola.
y = -x² + 4 x + 3
y = -(x² - 4 x - 3)
y = -{ x² - 2 (x) (2) + 2² - 2² - 3 }
y = - { (x - 2)² - 4 - 3 }
y = - { (x - 2)² - 7 }
y = - (x - 2)² + 7
y - 7 = -(x - 2)²
(y - k) = -4a (x - h)²
Here (h, k) is (2, 7) and the parabola is open downward.
Example 2 :
Find the maximum and minimum value of the function
2x3 + 3x2 - 36x + 1
Solution :
Let y = f(x) = 2x3 + 3x2 - 36x + 1
f'(x) = 2(3x2) + 3 (2x) - 36 (1) + 0
f'(x) = 6x2 + 6x - 36
set f'(x) = 0
6x² + 6x - 36 = 0
÷ by 6 => x² + x - 6 = 0
(x - 2)(x + 3) = 0
x - 2 = 0
x = 2
x + 3 = 0
x = -3
f'(x) = 6x² + 6x - 36
f''(x) = 6(2x) + 6(1) - 0
f''(x) = 12x + 6
Put x = 2
f''(2) = 12(2) + 6
= 24 + 6
f''(2) = 30 > 0 Minimum
To find the minimum value let us apply x = 2 in the original function
f(2) = 2(2)3 + 3(2)2 - 36(2) + 1
= 2(8) + 3(4) - 72 + 1
= 16 + 12 - 72 + 1
= 29 - 72
f(2) = -43
Put x = -3
f''(-3) = 12(-3) + 6
= -36 + 6
f''(-3) = -30 > 0 Maximum
To find the maximum value let us apply x = -3 in the original function
f(-3) = 2 (-3)3 + 3 (-3)2 - 36 (-3) + 1
= 2(-27) + 3(9) + 108 + 1
= -54 + 27 + 109
= -54 + 136
= 82
Step-by-step explanation: