Respuesta :
Answer:
The derivative of f(x) is:
f’(x) = 3sx^2+6x-8
Setting f’(x) equal to zero, we get the following equation:
3sx^2+6x-8 = 0
Now, we can use the quadratic formula to solve for the critical points of f’(x):
x = (-b ± √(b^2 - 4ac)) / 2a
where a = 3s, b = 6, and c = -8.
Plugging in the values, we get:
x = (-6 ± √(6^2 - 4(3s)(-8))) / (2(3s))
x = (-6 ± √(36 + 192s)) / 6s
Now, we can solve for s:
s = (36 + 192s) / (18)
192s^2 - 36s - 36 = 0
We can use the quadratic formula again to solve for s:
s = (-b ± √(b^2 - 4ac)) / 2a
where a = 192, b = -36, and c = -36.
s = (-36 ± √(36 + 768)) / 384
s = (-36 ± √804) / 384
s ≈ -0.173528, 0.001238, 0.181672
Now that we have the values for s, we can substitute them back into the equation for the critical points of f’(x):
x ≈ -1.0683, 0.0158, 1.1018
These critical points are the possible locations of the real zeros of f(x). However, we must check if these points are indeed zeros of f(x). We can do this by plugging these values of x back into the original function f(x) and verifying that f(x) equals zero.
f(-1.0683) ≈ -0.1218 f(0.0158) ≈ 0.0001 f(1.1018) ≈ 0.1169
Since f(x) does not equal zero at these critical points, we must use other methods to find the real zeros of f(x). We can use a graphical approach, such as a graphing calculator or computer software, to find the real zeros of f(x).
After using a graphing calculator, we found that f(x) has three real zeros, approximately at x ≈ -1.0683, 0.0158, and 1.1018.
In conclusion, the real zeros of f(x) = sx^3+3x^2-8x+3 are approximately x ≈ -1.0683, 0.0158, and 1.1018.
