Use method of calculus to find (a) any critical values and (b) any relative maxima of the function
h(x) = x^3 + 11/2 x^2 + 6x + 1
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To find the critical values of the function h(x), we need to find the values of x where the derivative of h(x) is equal to zero or does not exist.
(a) Finding the derivative of h(x):
h'(x) = 3x^2 + 11x + 6
Setting h'(x) = 0 and solving for x:
3x^2 + 11x + 6 = 0
Factorizing the quadratic equation:
(3x + 2)(x + 3) = 0
Setting each factor equal to zero:
3x + 2 = 0 ---> x = -2/3
x + 3 = 0 ---> x = -3
Therefore, the critical values of h(x) are x = -2/3 and x = -3.
(b) To determine if these critical values are relative maxima of the function h(x), we can analyze the second derivative of h(x).
Finding the second derivative of h(x):
h''(x) = 6x + 11
Substituting the critical values into h''(x):
h''(-2/3) = 6(-2/3) + 11 = -4 + 11 = 7
h''(-3) = 6(-3) + 11 = -18 + 11 = -7
Since h''(-2/3) > 0 and h''(-3) < 0, we can conclude that x = -2/3 is a relative minimum and x = -3 is a relative maximum of the function h(x).
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