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Chapter 1: Problem 53

Use a graphing utility to graph the function and approximate (to two decimal places) any relative minima or maxima. \(h(x)=(x-1) \sqrt{x}\)

Short Answer

Expert verified
The function \(h(x)=(x-1) \sqrt{x}\) has its minimum point at (1, 0).

Step by step solution

01

Find the derivative

The derivative of \(h(x)=(x-1) \sqrt{x}\) is found using the product rule. The derivative \(h'(x)\) will therefore be \( \frac{3}{2}\sqrt{x} - \frac{1}{2\sqrt{x}} \).
02

Find the critical values

The critical values of a differentiable function are those at which the derivative equals zero, and those where the derivative is undefined. In the case of \(h'(x)\), solve for \(x\) in \( \frac{3}{2}\sqrt{x} - \frac{1}{2\sqrt{x}} = 0 \). This gives the critical values at \(x= 0\) and \(x=1\). Note that \(x=0\) is not included in the domain given by the original function therefore the critical value to consider will be \(x=1\).
03

Apply the second derivative test

Find the second derivative \(h''(x)\), it would be \( \frac{3}{4\sqrt{x}} + \frac{1}{4x\sqrt{x}} \). For the second derivative test, look at the sign of the second derivative at the critical value. For \(x=1\), \(h''(1) = \frac{3}{4} + \frac{1}{4} = 1 \) which is positive, which means the function has a minimum point at \(x=1\).
04

Determine the minimum point

The minimum point of the function is at \( (1, h(1))\), therefore \( (1, 0) \) is the minimum point since \(h(x) = 0\) when \(x = 1\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Derivatives
In calculus, the derivative of a function measures the rate at which the function's value changes as its input changes. Think of it like the slope of a curve at any given point. This slope tells us how steep or flat the curve is, which is crucial for analyzing function behavior.
To find the derivative of a function like \(h(x)=(x-1) \sqrt{x}\), we often use the product rule. The product rule is handy when differentiating products of functions. For our example, it helps break down the expression into more manageable parts:
  • The first component: \((x-1)\)
  • The second component: \(\sqrt{x}\)
When you apply the product rule, the derivative\(h'(x)\) becomes \( \frac{3}{2}\sqrt{x} - \frac{1}{2\sqrt{x}} \).
This derivative tells us how the function \(h(x)\) behaves across its domain.
Finding Critical Values
Critical values are crucial points where a function's derivative is either zero or undefined. These points give us insight into where the function could experience changes in behavior, like peaks (maxima) or valleys (minima).
To find critical values for the derivative \(h'(x) = \frac{3}{2}\sqrt{x} - \frac{1}{2\sqrt{x}}\), we solve \(h'(x) = 0\) and consider where it may be undefined. By setting the equation to zero, you arrive at critical values for \(x=0\) and \(x=1\).
It's also essential to check the domain of the original function, \((x-1)\sqrt{x}\), to ensure all critical values make sense. In this problem, \(x=0\) is not within the domain, leaving \(x=1\) as the sole critical value of interest.
Applying the Second Derivative Test
The second derivative test helps determine whether a critical point is a relative minimum, maximum, or a point of inflection. This is done by examining the sign of the second derivative at the critical value.
For the function in question, we find the second derivative: \(h''(x) = \frac{3}{4\sqrt{x}} + \frac{1}{4x\sqrt{x}}\). By evaluating this second derivative at the critical value \(x=1\), we find \(h''(1) = \frac{3}{4} + \frac{1}{4} = 1\).
  • If the second derivative at a critical point is positive like here, the function is concave up, indicating a relative minimum.
  • If it were negative, it would indicate a relative maximum because the curve is concave down.
Thus, the point \((1,0)\) is a relative minimum for the function \(h(x)=(x-1) \sqrt{x}\).

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