Find the intervals of increase and decrease for the function h(x) = x^5 -2x^3 + x.

### 1 Answer | Add Yours

The function h(x) = x^5 -2x^3 + x

h'(x) = 5x^4 - 6x^2 + 1

The function is increasing intervals where the value of h'(x) > 0

5x^4 - 6x^2 + 1 > 0

=> (x-1)*(x+1)*(5*x^2-1) > 0

This is true when:

- (x-1) > 0, (x+1) > 0 and (5*x^2-1) > 0

=> x > 1, x > -1 and x^2 > 1/5

=> x > 1

- (x-1) < 0, (x+1) < 0 and (5*x^2-1) > 0

=> x < 1, x < -1 and x^2 > 1/5

=> x < -1

- (x-1) > 0, (x+1) < 0 and (5*x^2-1) < 0

=> x > 1, x < -1 and x^2 < 1/5

This is not possible for any value of x

- (x-1) < 0, (x+1) > 0 and (5*x^2-1) < 0

=> x < 1, x > -1 and x^2 < 1/5

=> `x in (-sqrt(1/5), sqrt(1/5))`

The function is decreasing in the interval `(-1, -sqrt(1/5))U(sqrt(1/5), 1)`

**The given function is increasing in `(-oo, -1)U(-sqrt(1/5), sqrt(1/5))U(1, oo)` and decreasing in **`(-1, -sqrt(1/5))U(sqrt(1/5), 1)`

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes