Question

Find the intervals in which the functions f(x)=x^3+5x^2-8x+1 is strictly increasing and decreasing.

Answer 1

Answer:

Step-by-step explanation:

Solution :

f(x)=x4−8x3+22x2−24x+21

f′(x)=4x3−24x+44x−24

To determine the intervals, f′(x)=0

4x3−24x2+44x−24=0

on factorizing

f′(x)=4(x−1)(x2−5x+6)

=4(x−1)(x−3)(x−2)

For f(x) to be increasing ,

f′(x)>0

4(x−1)(x2−5x+6)>0

=> 4(x−1)(x−3)(x−2)>0

=> 1<x<2 or 3<x<∞

=> x×(1,2)∪(3,8)

So f(x) is increasing on (1,2)∪(3,8)

For f(x) to be decreasing ,

f′(x)<0

=> (x−1)(x−3)(x−2)<0

=> 2<x<3 or x<1

=> x∈(2,3)∪(−∞,1)

So f(x) is decreasing on (2,3)∪(−∞,1)

Answer 2

f(x) = $x^3+5x^2-8x+1$ is strictly decreasing in the intervals (-∞, -4), (-4, 2/3) and (2/3, ∞).

Given,

f(x) = $x^3+5x^2-8x+1$

To find,

Find the intervals in which the functions f(x)=x^3+5x^2-8x+1 is strictly increasing and decreasing.

Solution,

f(x)=$x^3+5x^2-8x+1$

f′(x) = 3x² + 10x - 8

To determine the intervals, f′(x)=0

3x² + 10x - 8 = 0

On factorizing

f′(x) = 3x² + 12x - 2x - 8 =

      = 3x(x + 4) - 2(x + 4)

      = (3x-2)(x+4)

   x = 2/3 or -4

In the interval (-∞, -4), f(x) is negative

In the interval (-4, 2/3), f(x) is negative

In the interval (2/3, ∞), f(x) is negative

Hence, f(x) is strictly decreasing.

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