Answer the following questions for the function [tex]f(x)=x \sqrt{x^2+25}[/tex] defined on the interval [tex]-5 \leq x \leq 5[/tex].

[tex]f(x)[/tex] is concave down on the interval [tex]x=[/tex] $\square$ to [tex]x=[/tex] $\square$
[tex]f(x)[/tex] is concave up on the interval [tex]x=[/tex] $\square$ to [tex]x=[/tex] $\square$

The inflection point for this function is at [tex]x=[/tex] $\square$

The minimum for this function occurs at [tex]x=[/tex] $\square$

The maximum for this function occurs at [tex]x=[/tex] $\square$

Calculate the first derivative of f ( x ) = x x 2 + 25 ​ using the product and chain rules, resulting in f ′ ( x ) = x 2 + 25 ​ 2 x 2 + 25 ​ .
Compute the second derivative using the quotient rule, which simplifies to f ′′ ( x ) = ( x 2 + 25 ) 3/2 x ( 2 x 2 + 75 ) ​ .
Determine concavity by analyzing the sign of f ′′ ( x ) . The function is concave down on [ − 5 , 0 ) and concave up on ( 0 , 5 ] . The inflection point is at x = 0 .
Evaluate f ( x ) at the endpoints x = − 5 and x = 5 to find the minimum and maximum values. The minimum occurs at x = − 5 , and the maximum occurs at x = 5 .

f ( x ) is concave down on [ − 5 , 0 ) ​ f ( x ) is concave up on ( 0 , 5 ] ​ Inflection point at x = 0 ​ Minimum at x = − 5 ​ Maximum at x = 5 ​
Explanation

Problem Analysis We are given the function f ( x ) = x x 2 + 25 ​ defined on the interval − 5 ≤ x ≤ 5 . We need to find the intervals where f ( x ) is concave up and concave down, the inflection point, and the minimum and maximum values of f ( x ) on the given interval.

Finding the First Derivative First, we find the first derivative, f ′ ( x ) . Using the product rule and chain rule:


f ′ ( x ) = ( 1 ) ( x 2 + 25 ​ ) + x ( 2 1 ​ ( x 2 + 25 ) − 1/2 ( 2 x ))
f ′ ( x ) = x 2 + 25 ​ + x 2 + 25 ​ x 2 ​
f ′ ( x ) = x 2 + 25 ​ x 2 + 25 + x 2 ​
f ′ ( x ) = x 2 + 25 ​ 2 x 2 + 25 ​

Finding the Second Derivative Next, we find the second derivative, f ′′ ( x ) . Using the quotient rule:

f ′′ ( x ) = ( x 2 + 25 ​ ) 2 ( 4 x ) ( x 2 + 25 ​ ) − ( 2 x 2 + 25 ) ( 2 1 ​ ( x 2 + 25 ) − 1/2 ( 2 x )) ​
f ′′ ( x ) = ( x 2 + 25 ) 3/2 4 x ( x 2 + 25 ) − x ( 2 x 2 + 25 ) ​
f ′′ ( x ) = ( x 2 + 25 ) 3/2 4 x 3 + 100 x − 2 x 3 − 25 x ​
f ′′ ( x ) = ( x 2 + 25 ) 3/2 2 x 3 + 75 x ​
f ′′ ( x ) = ( x 2 + 25 ) 3/2 x ( 2 x 2 + 75 ) ​

Determining Concavity To find the intervals where f ( x ) is concave up and concave down, we analyze the sign of f ′′ ( x ) . The denominator ( x 2 + 25 ) 3/2 is always positive. The term 2 x 2 + 75 is also always positive. Therefore, the sign of f ′′ ( x ) depends on the sign of x .

0"> f ′′ ( x ) > 0 when 0"> x > 0 , so f ( x ) is concave up on the interval ( 0 , 5 ] .
f ′′ ( x ) < 0 when x < 0 , so f ( x ) is concave down on the interval [ − 5 , 0 ) .

Finding the Inflection Point To find the inflection point, we solve f ′′ ( x ) = 0 . This occurs when x = 0 .

Thus, the inflection point is at x = 0 .

Finding Minimum and Maximum Values To find the minimum and maximum values of f ( x ) on the interval [ − 5 , 5 ] , we evaluate f ( x ) at the endpoints and at any critical points. The critical points occur where f ′ ( x ) = 0 or is undefined. However, f ′ ( x ) = x 2 + 25 ​ 2 x 2 + 25 ​ is never 0 and is always defined. Thus, there are no critical points in the interior of the interval.

We evaluate f ( x ) at the endpoints:
f ( − 5 ) = − 5 ( − 5 ) 2 + 25 ​ = − 5 50 ​ = − 5 ( 5 2 ​ ) = − 25 2 ​
f ( 5 ) = 5 ( 5 ) 2 + 25 ​ = 5 50 ​ = 5 ( 5 2 ​ ) = 25 2 ​
Therefore, the minimum value occurs at x = − 5 and the maximum value occurs at x = 5 .

Final Answer Based on the calculations:

f ( x ) is concave down on the interval [ − 5 , 0 ) .
f ( x ) is concave up on the interval ( 0 , 5 ] .
The inflection point for this function is at x = 0 .
The minimum for this function occurs at x = − 5 .
The maximum for this function occurs at x = 5 .
Examples
Understanding concavity and inflection points is crucial in various real-world applications. For instance, when designing a bridge, engineers analyze the concavity of the structure to ensure stability and prevent collapse under different loads. Similarly, in economics, understanding the concavity of a cost function helps businesses optimize production levels to minimize costs and maximize profits. These concepts also play a vital role in physics, where analyzing the concavity of a potential energy function helps determine the stability of a system.

Answered by GinnyAnswer | 2025-07-05