Question #181816

Find the intervals of increase and decrease for the following functions: (a) f(x) = x 3 + 4x + 1

(b) f(x) = x 3 (5 − x) 2 .

(c) f(x) = x + sin x.

(d) f(x) = (x 2 − 4).

(e) f(x) = 2x 3 − 9x 2 + 12x

(f) f(x) = (x 2 − 4)2

(g) f(x) = x 3 − 2x 2 + x + 5 


1
Expert's answer
2021-05-07T14:33:30-0400

To find the intervals of increase/decrease of f(x), we need to find intervals, where its derivative is positive and negative respectively.

(a) f(x)=x3+4x+1,f(x)=3x2+44xRf(x) = x^3 +4x + 1, f'(x) = 3x^2+4 \geqslant 4\:\:\forall x \in \mathbb{R} , so f(x) increases at whole real number axis

(b) f(x)=x3(5x)2f(x) = x^3(5-x)^2

f(x)=3x2(5x)2+x32(5x)(1)==x2(x5)(3(x5)+2x)=x2(x5)5(x3)f'(x) = 3x^2(5-x)^2 + x^3 \cdot2(5-x) \cdot (-1) = \\ = x^2(x-5) \cdot (3(x-5)+2x) = x^2(x-5)5(x-3) , and f'(x) sign is the same that sign of (x3)(x5):(x-3)(x-5) : it's <0< 0 (f(x) decreases) for x(3,5)x \in (3, 5) , it's >0>0 (f(x) increases) for x(,3)(5,+)x \in (-\infty, 3) \cup(5, +\infty)

(c) f(x)=x+sinx,f(x)=1+cosx0xRf(x) = x+\text{sin}x, f'(x) = 1+\text{cos}x \geqslant 0 \:\: \forall x \in \mathbb{R} , so f(x) increases in all points except f'(x) nulls : {xR:cosx=1}={π+2πn,nZ}\{x \in \mathbb{R}: \text{cos}x = -1\} = \{\pi+2\pi n, n \in \mathbb{Z}\}

(d) f(x)=x24,f(x)=2xf(x) = x^2-4, f'(x) = 2x , and x(,0)f(x)<0\forall x \in (-\infty, 0) \: f'(x)<0 (f(x) decreases), andx(0,+)f(x)>0\forall x \in (0, +\infty) \: f'(x)>0 (f(x) increases)

(e) f(x)=2x39x2+12x,f(x)=6x218x+12f(x) = 2x^3-9x^2+12x, f'(x) = 6x^2-18x+12 , so f(x)=6(x23x+2)=6(x1)(x2)f'(x) = 6(x^2-3x+2) = 6(x-1)(x-2) and x(1,2)f(x)<0\forall x \in (1, 2) \: f'(x)<0 (f(x) decreases), and x(,1)(2,+)f(x)>0\forall x \in (-\infty, 1)\cup(2, +\infty) \: f'(x)>0 (f(x) increases)

(f) f(x)=(x24)2,f(x)=2(x24)2xf(x) = (x^2-4)^2, f'(x) = 2(x^2-4)\cdot2x , f(x)=4(x+2)x(x2)f'(x) = 4(x+2)x(x-2) , so x(,2)(0,2)f(x)<0\forall x \in (-\infty, -2)\cup(0, 2) \: f'(x)<0 (f(x) decreases), x(2,0)(2,+)f(x)>0\forall x \in (-2, 0)\cup(2, +\infty) \: f'(x)>0 (f(x) increases)

(g) f(x)=x32x2+x+5,f(x)=3x24x+1f(x) = x^3-2x^2+x+5, f'(x) = 3x^2-4x+1 , so f(x)=(3x1)(x1)f'(x) = (3x-1)(x-1) and x(13,1)f(x)<0\forall x \in (\frac{1}{3}, 1) \: f'(x)<0 (f(x) decreases), and x(,13)(1,+)f(x)>0\forall x \in (-\infty, \frac{1}{3})\cup(1, +\infty) \: f'(x)>0(f(x) increases)


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS