Question

Determine all the points of local maxima and local minima of the following function: f(x) = (-¾)x4 – 8x3– (45/2)x2 + 105​

Answer 1

Answer:

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Explanation:

f (x) =  

= – 3x3 – 24x2 – 45x

= –3x (x2 + 8x + 15)

For local maximum or local minimum, we must have

= 0

– 3x (x2 + 8x + 15) = 0

– 3x (x + 3) (x + 5) = 0 x = 0, –3, –5

Thus, x = 0, x = –3 and x = –5 are the possible points of local maxima or minima.

For x=0, f'(x) changes from positive to negative. So x=0 is the point of local maxima.

For x=-3, f'(x) changes its sign from negative to positive. So x=-3 is a point of local minima.

For x=-5, f'(x) changes sign from positive to negative. So x=-5 is a point of local maxima.

Corresponding values of f(x), we get

x=0, f(x) = 105

x=-3, f(x) = 57.75

x=-5, f(x) = 73.75

Answer 2

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