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Chapter 4: Problem 62

Determine the intervals on which the following functions are concave up or concave down. Identify any inflection points. $$f(x)=2 x^{2} \ln x-5 x^{2}$$

Short Answer

Expert verified
Question: Determine the intervals on which the given function is concave up or concave down, and identify any inflection points for the function \(f(x)=2x^2\ln x -5x^2\). Answer: The function is concave up on the interval \((1, \infty)\), concave down on the interval \((0, 1)\), and has an inflection point at \((1, -5)\).

Step by step solution

01

Find the first derivative of the function

To begin, we need to find the first derivative of the given function: $$f(x)=2x^2\ln x -5x^2$$ Using the product and power rules, we have: $$f'(x)=2(2x\ln x + x^2(\frac{1}{x})) - 10x$$ Simplify the expression: $$f'(x)=4x\ln x + 2x^2 - 10x$$
02

Find the second derivative of the function

Next, we need to find the second derivative of the function by taking the derivative of \(f'(x)\): $$f''(x)=4(\ln x + 1) + 4x - 10$$
03

Find critical points of the second derivative

To determine the concavity of the function and find possible inflection points, we need to look for the critical points of the second derivative, which are the points where \(f''(x)=0\) or is undefined. In this case, \(f''(x)\) is defined for all x, so we only need to find the points where it is equal to 0: $$4(\ln x + 1) + 4x - 10 = 0$$ This equation cannot be easily solved analytically for x. Thus, we must consider the sign change of the second derivative to determine concavity.
04

Determine the concavity

Since we cannot solve for x explicitly, we will determine the intervals by analyzing the sign of the second derivative: 1. When \(x>1\), \(\ln x>0\), so \(f''(x)>0\). This means the function is concave up when \(x>1\). 2. When \(0<x<1\), \(\ln x<0\), so \(f''(x)<0\). This means the function is concave down when \(0<x<1\). Since \(f''(x)\) is not defined at \(x=0\), we do not need to analyze the concavity at \(x=0\)
05

Identify inflection points

Inflection points occur when the second derivative changes sign. In this case, the second derivative changes sign at \(x=1\). Therefore, there is an inflection point at \(x=1\). To find the corresponding y-value, plug \(x=1\) back into the original function: $$f(1) = 2(1)^2 \ln 1 - 5(1)^2 = 0 - 5 = -5$$ So, there is an inflection point at \((1, -5)\). In conclusion, the function is concave up on the interval \((1, \infty)\), concave down on the interval \((0, 1)\), and has an inflection point at \((1, -5)\).

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