Dear Student
Given function: f(x) = (-¾)x^4– 8x^3– (45/2)x^2+ 105
Thus, differentiate the function with respect to x, we get
f ′ (x) = –3x^3– 24x^2– 45x
Now take, -3x as common:
= – 3x (x^2+ 8x + 15)
Factorise the expression inside the bracket, then we have:
= – 3x (x +5)(x+3)
f ′ (x) = 0
⇒ x = –5, x = –3, x = 0
Now, again differentiate the function:
f ″(x) = –9x^2– 48x – 45
Take -3 outside,
= –3 (3x^2+ 16x + 15)
Now, substitue the value of x in the second derivative function.
f ″(0) = – 45 < 0. Hence, x = 0 is point of local maxima
f ″(–3) = 18 > 0. Hence, x = –3 is point of local minima
f ″(–5) = –30 < 0. Hence, x = –5 is point of local maxima.
Thanks