f(x) = -x² + 4x + 3|
\(\frac{d}{dx}[-x^{2}] + \frac{d}{dx}[4x] + \frac{d}{dx}[3]|\)
\(-2x + 4 + 0|\)
\(-2x + 4|\)
\(\frac{d}{dx}[x^{n}]| = nx^{n - 1}|\)
\(\frac{d}{dx}[-x^{2}]| = 2(-x)^{2 - 1}|\)
\(\frac{d}{dx}[-x^{2}| = 2(-x)^{1}|\)
\(\frac{d}{dx}[-x^{2}]| = 2(-x)|\)
\(\frac{d}{dx}[-x^{2}]| = -2x|\)
\(\frac{d}{dx}[4x]|\)
\(\frac{4dx}{dx}|\)
\(4(1)|\)
\(4|\)
\(\frac{d}{dx}[3]|\)
\(\frac{3d}{dx}|\)
\(3(1)(0)|\)
\(3(0)|\)
\(0|\)
-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.
