Answer the following questions for the function
f(x)=x(sqrtx2+25)
defined on the interval [–5,4].
A. f(x) is concave down on the region to
B. f(x) is concave up on the region to
C. The inflection point for this function is at
D. The minimum for this function occurs at
E. The maximum for this function occurs at
assuming you know how to use derivatives to answer the questions,
f = x√(x^2+25)
f' = (2x^2+25)/√(x^2+25)
f" = x(2x^2+75)/(x^2+25)^(3/2)
That should help. Show where you get stuck, if you do.
I tried putting the first derivative =0 so i can find my max and mins but for some reason im having trouble getting the right answer it comes out as a complex number.. please help!?
To determine the concavity of the function \(f(x)\), we need to find the second derivative of the function and evaluate it.
1. First, let's find the first derivative of \(f(x)\):
\(f'(x) = \frac{d}{dx}\left[x(\sqrt{x^2 + 25})\right]\)
Using the product rule, we can differentiate \(x\) and \(\sqrt{x^2 + 25}\) separately:
\(f'(x) = \sqrt{x^2+25} + \frac{x}{2\sqrt{x^2+25}}(2x)\)
\(f'(x) = \sqrt{x^2+25} + \frac{x^2}{\sqrt{x^2 + 25}}\)
2. Now, let's find the second derivative of \(f(x)\):
\(f''(x) = \frac{d}{dx}\left(\sqrt{x^2+25} + \frac{x^2}{\sqrt{x^2 + 25}}\right)\)
To differentiate \(\sqrt{x^2+25}\), we use the chain rule:
\(f''(x) = \frac{1}{2\sqrt{x^2+25}}(2x) + \frac{d}{dx}\left(\frac{x^2}{\sqrt{x^2 + 25}}\right)\)
The derivative of \(\frac{x^2}{\sqrt{x^2 + 25}}\) involves the quotient rule:
\(f''(x) = \frac{x}{\sqrt{x^2+25}} + \frac{\left(\frac{x^2}{\sqrt{x^2 + 25}}\right)(2x)}{x^2+25}\)
Simplifying further:
\(f''(x) = \frac{x}{\sqrt{x^2+25}} + \frac{2x^3}{(x^2+25)\sqrt{x^2 + 25}}\)
3. To find the concavity, we need to evaluate \(f''(x)\) at the critical points within the interval \([-5, 4]\).
Let's find the critical points by setting the first derivative equal to zero:
\(\sqrt{x^2+25} + \frac{x^2}{\sqrt{x^2 + 25}} = 0\)
Multiplying through by \(\sqrt{x^2 + 25}\):
\(x^2 + 25 + x^2 = 0\)
Simplifying:
\(2x^2 + 25 = 0\)
Solving for \(x\):
\(x^2 = -\frac{25}{2}\)
Since \(x\) must be real, there are no critical points within the interval \([-5, 4]\).
4. Next, we need to determine the inflection point by finding where the concavity changes. This occurs when the second derivative equals zero or is undefined.
Setting \(f''(x)\) equal to zero:
\(\frac{x}{\sqrt{x^2+25}} + \frac{2x^3}{(x^2+25)\sqrt{x^2 + 25}} = 0\)
Multiplying through by \((x^2 + 25)\sqrt{x^2 + 25}\):
\(x(x^2 + 25) + 2x^3 = 0\)
Simplifying:
\(x^3 + 25x + 2x^3 = 0\)
Combining like terms:
\(3x^3 + 25x = 0\)
Factoring out \(x\):
\(x(3x^2 + 25) = 0\)
Therefore, the inflection point occurs when \(x = 0\).
5. We need to find the minimum and maximum points by determining the behavior of the function at the endpoints of the interval \([-5, 4]\) since there are no critical points within this interval.
Evaluating the function at the endpoints:
\(f(-5) = -5\sqrt{(-5)^2 + 25} = -5\sqrt{50} = -5\sqrt{25 \cdot 2} = -5 \cdot 5 \sqrt{2} = -25\sqrt{2}\)
\(f(4) = 4\sqrt{4^2 + 25} = 4\sqrt{16 + 25} = 4\sqrt{41}\)
Therefore, the minimum point occurs at \(x = -5\) with a function value of \(-25\sqrt{2}\), and the maximum point occurs at \(x = 4\) with a function value of \(4\sqrt{41}\).
Finally, let's answer each question:
A. As there are no critical points, we cannot determine where \(f(x)\) is concave down. The answer to this question would be "unknown".
B. Similarly, we cannot determine where \(f(x)\) is concave up. The answer to this question would also be "unknown".
C. The inflection point for this function is at \(x = 0\).
D. The minimum for this function occurs at \(x = -5\), where \(f(x) = -25\sqrt{2}\).
E. The maximum for this function occurs at \(x = 4\), where \(f(x) = 4\sqrt{41}\).