calc! help!! thank you ^^?

Question

find the extreme values of the function and where they occur

y= x^3 -2x +4

Answer 1

To find extreme value, need to find maximum or minimum values

y = x^3-2x+4
y' = 3x^2 - 2
y" = 6x

To find extreme values,
y' = 3x^2 - 2 = 0
x^2 = 2/3
x = 0.816, -0.816

When x = 0.816, y = (0.816)^3-2(0.816)+4 = 2.91
y" = +positive (minimum value)

When x = -0.816, y = (-0.816)^3-2(-0.816)+4 = 5.09
y" = -negative(maximum value)

So, minimum value = 2.91 at x = 0.816
maximum value = 5.09 at x = -0.816

Answer 2

Extreme values is a misnomer for relative, or local minima and maxima. The extremes of this function cannot be calculated. For local minimum and/or maximum,
y' = 3x^2 - 2 = 0
x = ± √(2/3)
y(-√(2/3)) = -(√(2/3))^3 + 2√(2/3) + 4 = 4.9623

y(√(2/3)) = (√(2/3))^3 - 2√(2/3) + 4 = 3.0377

Answer 3

I assume you mean the relative maximum and relative minimum.

f(x)= x³ - 2x + 4

f'(x) = 3x² - 2 = 0
x² = 2/3
x = ±√(2/3)

The second derivative tells us what nature of the critical values:

f''(x) = 6x
6√(2/3) > 0 implies relative minimum at x = +√(2/3) ≈ 0.8164965
-6√(2/3) < 0 implies relative maximum at x = -√(2/3) ≈ -0.8164965


f(x) = x³ - 2x + 4

f(√(2/3)) = (2/3)^(3/2) - 2√(2/3) + 4 ≈ 2.9113379
f(-√(2/3)) = -(2/3)^(3/2) + 2√(2/3) + 4 ≈ 5.0886621

Answer 4

hmmm
y'= 3x^2-2