1) The graph of the function f(x)= 4x^3 - 48x has one local maximum and one loca

Question

1) The graph of the function f(x)= 4x^3 - 48x has one local maximum and one local minimum point. Find these points using the first derivative test.

2) The graph of the following function has one relative maximum point and one relative minimum point. Find these points using the first derivative test. f(x) = x^3 + 9x^2 + 15x + 1

3) The graph of the following function has one relative maximum point and one relative minimum point. Find these points using the first derivative test. f(x) = x^3 + 3x^2 + 2

Explanation / Answer

(1) f(x) = 4x^3 - 48x

f'(x) = 12x^2 - 48

f'(x) = 0

12x^2 - 48 = 0 which gives x = {-2, 2}

f'(-2.1) = 12(-2.1)^2 - 48 = 4.92 and f'(-1.9) = 12(-1.9)^2 - 48 = -4.68

So x = -2 is a point of local maximum and the maximum value is f(-2) = 4(-2)^3 - 48(-2) = 64

So the extreme point is (-2, 64)

f'(1.9) = 12(1.9)^2 - 48 = -4.68 and f'(2.1) = 12(2.1)^2 - 48 = 4.92

So x = 2 is a point of local minimum and the minimum value is f(2) = 4(2)^3 - 48(2) = -64

So the extreme point is (2, -64)

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