Question

Number of points of local maxima and minima of the function
$f(x) = - \dfrac{3}{4} {x}^{4} - 8 {x}^{3} - \dfrac{45}{2} {x}^{2} + 105 \: are$
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Answer 1

we have to find the number of points of local maxima and minina of given function, f(x) = -3/4 x⁴ - 8x³ - 45/2 x² + 105

solution : To find local maximum and minimum

if y = f(x) is a function where f''(x) > 0 at x = a , at x = a, will be minimum

while f"(x) < 0, at x = b then at x = b, will be maximum.

where at f"(x) = 0, x = a , b

let's solve it using above concept.

f'(x) = -3 x³ - 24x² - 45x

= -3x(x² + 8x + 15)

= -3x(x + 3)(x + 5)

at f'(x) = 0, x = 0, -3, - 5

now differentiating once more w.r.t x,

f"(x) = - 9x² - 48x - 45

at x = 0, f"(x) < 0, local maxima

at x = -3 , f"(x) > 0 , local minima

at x = -5 , f"(x) < 0 , local maxima

hence there are three points where function gets local maxima and minima.

Answer 2

Solution :

f(x)= -3x³ - 24x² - 45x

= -3x(x² + 8x + 15)

= -3x(x + 5) (x + 3)

f(x) = 0

x = -5, x = -3,x = 0

f(x)=-9x² - 48x - 45

= - 3(3x² + 16x + 15)

f(0)= -45 < 0.

Therefore, x=0 is point of local maxima

f(-3) = 18 > 0.

Therefore, x=-3 is point of local minima

f(-5)= -30 < 0.

Therefore, x=-5 is point of local maxima

Extra information :

(a² + b²) = a² + b² + 2ab

(a-b)² = a² + b² - 2aba² - b² = (a + b)(a - b)

a²+b²-(a+ b)²- 2ab or a² +b²(a - b)² +2aba³ +b³ = (a+ b)(a²- ab + b) = (a +b)³ - 3ab(a + b)

a³-b³ = (a- b)(a² + ab + b) = (a-b)³+ 3ab(a-b)