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Chapter 7: Problem 67

If \(f(x)=-\frac{1}{3} x^{3}+t^{2} x\), where \(t\) is a real parameter. Let \(m(t)\) denote the minimum of \(\mathrm{f}(\mathrm{x})\) over \([0,1]\) then (A) \(m(t)=0\) if \(t^{2} \geq \frac{1}{3}\) (B) \(\mathrm{m}(\mathrm{t})=0\) for all \(\mathrm{t}\) (C) \(m(t)=t^{2}-\frac{1}{3}\) if \(t^{2}<\frac{1}{3}\) (D) \(\mathrm{m}(\mathrm{t})=\frac{1}{3}-\mathrm{t}^{2}\) for all \(\mathrm{t}\).

Short Answer

Expert verified
a) m(t) = 0 if t^2 >= 1/3 c) m(t) = -8/3*(t^2)^3 + 6/3*(t^2)^2 if t^2 < 1/3 The correct options are (A) and (C).

Step by step solution

01

Find the derivative of the function

We need to differentiate the function f(x) to find its critical points. The derivative of f(x) is given by: f'(x) = -x^2+(2t^2) x
02

Find the critical points

Critical points occur where f'(x)=0 or f'(x) is undefined. Here, f'(x) is a quadratic function and is always defined. To find where f'(x)=0, we solve the equation: -x^2+(2t^2) x = 0 This simplifies to x(2t^2-x)=0. So, the critical points occur at x=0 and x=2t^2.
03

Determine whether the critical points are in the interval [0, 1]

Since x=0 is the left endpoint of the interval, we need to determine if x=2t^2 is in the interval [0, 1]. For this to happen, 0 鈮� 2t^2 鈮� 1. Dividing by 2, we get 0 鈮� t^2 鈮� 1/2.
04

Determine the minimum value of f(x) on the interval [0, 1]

Our possible minimum values occur at x=0, and x=2t^2 (when 0 鈮� t^2 鈮� 1/2). We will evaluate f(x) at these points and compare their values. For x=0, f(x) = t^2 * 0 = 0 For x=2t^2, f(x) = -\frac{1}{3} (2t^2)^3 + t^2(2t^2) = -\frac{8}{3}(t^2)^3 + 2(t^2)^2 = -\frac{8}{3}(t^2)^3 + \frac{6}{3}(t^2)^2 Now we need to compare f(0) and f(2t^2) to find the minimum m(t) on the interval [0,1].
05

Compare the function values and choose the correct option

We have: f(0) = 0 f(2t^2) = -\frac{8}{3}(t^2)^3 + \frac{6}{3}(t^2)^2 If t^2 < \frac{1}{3}, f(2t^2) = -\frac{8}{3}(t^2)^3 + \frac{6}{3}(t^2)^2, which is negative since the first term is larger than the second term. Thus, m(t) = f(2t^2) in this case. If t^2 鈮� \frac{1}{3}, f(2t^2) 鈮� 0, so f(0) would be the minimum value on the interval. Thus, m(t) = 0 in this case. Comparing our findings to the given options, we can conclude that: (A) m(t) = 0 if t^2 鈮� \frac{1}{3} is correct. (C) m(t) = t^2 - \frac{1}{3} if t^2 < \frac{1}{3} is also correct. The correct options are (A) and (C).

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