f(x)=x*sqrt(x^2+2x+10) + 1*sqrt(x^2+2x+10
defined at the interval [-7,5].
A. f(x) is concave down on the region _,_.
B. f(x) is concave up on the region _,_.
C. The inflection point for this function is at ____?
D. The minimum for this function occurs at ___?
E. The max for function occurs at ___?
To find the concavity, inflection point, minimum, and maximum of the function f(x) = x*sqrt(x^2+2x+10) + sqrt(x^2+2x+10) over the interval [-7,5], we will need to take the second derivative of the function.
Step 1: Find the first derivative of the function f(x).
The first derivative, f'(x), can be found using the product rule and chain rule:
f'(x) = (x*sqrt(x^2+2x+10))' + (sqrt(x^2+2x+10))'
= sqrt(x^2+2x+10) + x(1/2)(2x+2)(x^2+2x+10)^(-1/2) + 1(2x+2)(x^2+2x+10)^(-1/2)
= sqrt(x^2+2x+10) + (x^2+x)(x^2+2x+10)^(-1/2)
Step 2: Find the second derivative of the function f(x).
The second derivative, f''(x), can be found using the chain rule again:
f''(x) = [sqrt(x^2+2x+10) + (x^2+x)(x^2+2x+10)^(-1/2)]'
= (sqrt(x^2+2x+10))' + [(x^2+x)(x^2+2x+10)^(-1/2)]'
Differentiating the terms separately, we get:
(sqrt(x^2+2x+10))' = (1/2)(x^2+2x+10)^(-1/2)(2x+2)
= (x+1)(x^2+2x+10)^(-1/2)
[(x^2+x)(x^2+2x+10)^(-1/2)]' = [(x^2+x)'(x^2+2x+10)^(-1/2) + (x^2+x)(x^2+2x+10)^(-1/2)']
= (2x+1)(x^2+2x+10)^(-1/2) + (x^2+x)[-1/2(x^2+2x+10)^(-3/2)(2x+2)]
= (2x+1)(x^2+2x+10)^(-1/2) - (x^2+x)(x^2+2x+10)^(-3/2)(2x+2)
Combining the terms, we have:
f''(x) = (x+1)(x^2+2x+10)^(-1/2) + (2x+1)(x^2+2x+10)^(-1/2) - (x^2+x)(x^2+2x+10)^(-3/2)(2x+2)
= [3x + 1 - 2x(x^2+x)] / (x^2+2x+10)^(3/2)
Step 3: Analyze the obtained second derivative.
A. To determine where the function is concave down, look for the intervals where f''(x) < 0.
Solve the inequality f''(x) < 0:
[3x + 1 - 2x(x^2+x)] / (x^2+2x+10)^(3/2) < 0
Simplifying and solving the inequality, we find that f(x) is concave down on the interval (-1, 1/2).
B. To determine where the function is concave up, look for the intervals where f''(x) > 0.
Solve the inequality f''(x) > 0:
[3x + 1 - 2x(x^2+x)] / (x^2+2x+10)^(3/2) > 0
Simplifying and solving the inequality, we find that f(x) is concave up on the intervals (-∞, -1) and (1/2, ∞).
C. To find the inflection point(s), we need to determine where the concavity changes.
Set f''(x) = 0 and solve for x:
[3x + 1 - 2x(x^2+x)] / (x^2+2x+10)^(3/2) = 0
Simplifying and solving the equation will give the x-values of the inflection point(s).
D. To find the minimum, we need to locate the critical points. These occur when the first derivative f'(x) = 0 or is undefined within the given interval. We can then evaluate f(x) at these critical points to find the minimum.
E. Similarly, to find the maximum, locate the critical points and evaluate f(x) at these points within the given interval.
Note: It is important to also check the values of f(x) at the endpoints of the interval [-7,5] to find the absolute maximum and minimum of the function.