Given the function f(x)= x√(x+3),
Indicate the domain of f(x): …………………………
What are the x – intercepts of the graph of f(x)? POINTS! …………………………….
Identify the critical points of f(x) (write pairs of coordinates, not just x values)
The critical points are ………………………………………………….
Use the second derivative to determine whether each of the points found in part (c) is a local maximum or minimum
…………………………………………………………………………………………………………………………………
Use the first/second derivatives to describe the behavior of f(x) in the interval (-3,-2)
f(x) is ……………………………………….. (increasing/decreasing) on (-3, -2)
f(x) is concave …………………………. (up/down) on (-3, -2)
my professor never went over how to do this but so far I got the domain of x> or equal to 0/ x intercepts are (0,0) (-3,0)
f'(x) = sqrt(x+3) = x/(xsqrt(x+3))
with crit points of (-2,-2) and (-3,0)
stuck on d and e though
To determine the critical points of the function f(x), you need to find the values of x for which the derivative of f(x) equals 0 or is undefined.
The derivative of f(x) can be found using the power rule and the chain rule. Let's calculate it step by step:
f'(x) = (x√(x+3))'
Using the chain rule, the derivative of √(x+3) is 1/2√(x+3) times the derivative of (x+3):
f'(x) = (x√(x+3))' = √(x+3) + x(1/2)(x+3)^(-1/2)(1)
Simplifying further:
f'(x) = √(x+3) + x/(2√(x+3))
To find the critical points, we need to solve f'(x) = 0:
√(x+3) + x/(2√(x+3)) = 0
Multiplying both sides by 2√(x+3) to get rid of the denominator:
2(x+3) + x = 0
Expanding and simplifying the equation:
2x + 6 + x = 0
3x + 6 = 0
3x = -6
x = -2
Therefore, the critical point is (-2, f(-2)).
Now, let's analyze the behavior of f(x) using the second derivative.
The second derivative of f(x) can be found by differentiating f'(x):
f''(x) = (√(x+3) + x/(2√(x+3)))'
Using the quotient rule and the chain rule:
f''(x) = (1/(2√(x+3))) + (2(x+3)^(-1/2)(1/2))
Simplifying further:
f''(x) = 1/(2√(x+3)) + 1/(x+3)
To determine whether the critical point (-2, f(-2)) is a local maximum or minimum, we need to evaluate the second derivative at that point:
f''(-2) = 1/(2√(-2+3)) + 1/(-2+3)
f''(-2) = 1/(2√1) + 1/1
f''(-2) = 1/2 + 1
f''(-2) = 3/2
Since the second derivative is positive at x = -2, the critical point (-2, f(-2)) is a local minimum.
To describe the behavior of f(x) in the interval (-3, -2), we need to analyze the first derivative.
From part c, we found that the critical point (-2, f(-2)) is a local minimum.
To determine if f(x) is increasing or decreasing on (-3, -2), we can examine the sign of the first derivative f'(x) in that interval.
In part b, we found that f'(x) = √(x+3) + x/(2√(x+3)).
Now, evaluate f'(-3) and f'(-2):
f'(-3) = √(-3+3) - 3/(2√(-3+3)) = 0
f'(-2) = √(-2+3) - 2/(2√(-2+3)) = 0
Since both f'(-3) and f'(-2) are equal to 0, f(x) is neither increasing nor decreasing on the interval (-3, -2).
To determine the concavity of the function on (-3, -2), we need to analyze the second derivative f''(x).
In part d, we found that f''(x) = 1/(2√(x+3)) + 1/(x+3).
Now, evaluate f''(-3) and f''(-2):
f''(-3) = 1/(2√(-3+3)) + 1/(-3+3) = undefined
f''(-2) = 1/(2√(-2+3)) + 1/(-2+3) = ∞
Since f''(-3) is undefined and f''(-2) is infinite, we cannot determine the concavity of f(x) on the interval (-3, -2).
Therefore, we cannot conclude whether f(x) is concave up or concave down on (-3, -2).