Determine between which consecutive integers the real zeros of each function are located on the given interval.
5. f(x) = x^3 + 5x^2 – 4; [–6, 2]
Answers
when we take out roots of the following equation we are getting
x belongs from (-5 - √73)/6 to (-5 + √73)/6
Value of the equation will become 0 only on thesw values of x
these roots are real but they are not integer = Ans = 0 (or null value m not sure)
Given : f(x) = x³ + 5x² - 4
To find : Determine between which integers the real zeros of each function are located on the given interval
Solution:
f(x) = x³ + 5x² - 4
f(-1) = - 1 + 5 - 4 = 0
-1 is one the zero
=> x + 1 is one of the factor
-1 lies between [–6, 2]
x² + 4x - 4
x + 1 _| x³ + 5x² - 4 |_
x³ + x²
______
4x² - 4
4x² + 4x
_________
-4x - 4
-4x - 4
______
0
x³ + 5x² - 4 = (x + 1)(x² + 4x - 4)
x² + 4x - 4
x = (- 4 ± √16 + 16)/2
= -2 ± 2√2
= - 2 ± 2.83
= 0.83 , - 4.83
Zeroes are
- 4.83 , - 1, 0.83
=> these lies between integers - 5 & 1 which are with in [–6, 2]
real zeros of function x³ + 5x² - 4 are located between integers - 5 & 1
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