Question

Find the interval of decreasing y=-x^2+4x+3

Answer 1

f(x) = -x² + 4x + 3|

`(d)/(dx)[-x^(2)] + (d)/(dx)[4x] + (d)/(dx)[3]|`

`-2x + 4 + 0|`

`-2x + 4|`


`(d)/(dx)[x^(n)]| = nx^(n - 1)|`

`(d)/(dx)[-x^(2)]| = 2(-x)^(2 - 1)|`

`(d)/(dx)[-x^(2)| = 2(-x)^(1)|`

`(d)/(dx)[-x^(2)]| = 2(-x)|`

`(d)/(dx)[-x^(2)]| = -2x|`


`(d)/(dx)[4x]|`

`(4dx)/(dx)|`

`4(1)|`

`4|`


`(d)/(dx)[3]|`

`(3d)/(dx)|`

`3(1)(0)|`

`3(0)|`

`0|`

-2x + 4 = 0|
- 4 - 4
-2x = -4|
-2 -2
x = 2|

(-∞, 2) ∨ (2, ∞)|
(-∞, 2)| ∨| (2, ∞)|

(-∞, 2)
f'(x) = -2x + 4|
f'(2) = -2(2) + 4|
f'(2) = -4 + 4|
f'(2) = 0|

At 2| the derivative is 0.| Since this is positive, the function is increasing on (-∞, 2).
Increasing On: (-∞,| 2)| since f(x)| > 0|

(2, ∞)
f'(x) = -2x + 4|
f'(2) = -2(2) + 4|
f'(2) = -4 + 4|
f'(2) = 0|

At 2| the derivative is 0.| Since this is negative, the function is decreasing on (2, ∞)|.
Decreasing On: (2,| ∞)| since f(x)| < 0|

THe interval of the decreasing function y = -x² + 4x + 3 is (2, 2), or 2 > x > 2 and 2 < x < 2.