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Chapter 7: Problem 51

If $f(x)=\left\\{\begin{array}{ll}3 x^{2}+12 x-1 ;-1 \leq x \leq 2 \\ 37-x & ; 2

Short Answer

Expert verified
(A) The function \(f(x)\) is increasing in \([-1,2]\). (B) The function \(f(x)\) is continuous in \([-1,3]\). (C) The first derivative of \(f(x)\) does not exist at \(x=2\). (D) The function \(f(x)\) has the maximum value at \(x=2\). Answer: Options (A), (C), and (D) are true, while option (B) is false.

Step by step solution

01

Check if \(f(x)\) is increasing in \([-1, 2]\)

We are given that \(f(x) = 3x^2 + 12x - 1\) for \(-1 \leq x \leq 2\). Let's find the derivative of \(f(x)\) within this range. \(f'(x) = \frac{d}{dx}(3x^2 + 12x - 1) = 6x + 12\). Now, let us check whether the derivative is positive within the interval \([-1, 2]\): \(f'(-1) = 6(-1) + 12 = 6\) \(f'(2) = 6(2) + 12 = 24\) Since \(f'(-1)\) and \(f'(2)\) are both positive, we can safely assume that \(f'(x) > 0\) for all \(x \in [-1, 2]\). Hence, \(f(x)\) is increasing in \([-1,2]\). So, option (A) is true.
02

Check if \(f(x)\) is continuous in \([-1, 3]\)

The function is composed of a quadratic and a linear function, both of which are continuous functions on their respective domains. Thus, to verify if \(f(x)\) is continuous in \([-1, 3]\), we only need to check the continuity at the point of change- \(x=2\). \(f(2)_{left} = 3(2)^2 + 12(2) - 1 = 37\) \(f(2)_{right} = 37 - 2 = 35\) Since \(f(2)_{left}\) and \(f(2)_{right}\) differ, the function is not continuous in \([-1,3]\). So, option (B) is false.
03

Check if \(f'(2)\) exists

To determine if \(f'(2)\) exists, we should check if the function is differentiable at \(x=2\). Let's find the derivative of \(f(x)\) for \(2 < x \leq 3\), and then check for differentiability at \(x=2\). For \(2 < x \leq 3\), \(f(x) = 37-x\). So, \(f'(x) = -1\). Now, let's calculate the left- and right-hand derivatives of \(f(x)\) at \(x=2\): \(f'(2)_{left} = 6(2) + 12 = 24\) \(f'(2)_{right} = -1\) Since \(f'(2)_{left}\) and \(f'(2)_{right}\) differ, the function is not differentiable at \(x=2\), and \(f'(2)\) does not exist. So, option (C) is true.
04

Check if \(f(x)\) has the maximum value at \(x=2\)

To find out whether \(f(x)\) has the maximum value at \(x=2\), we should compare the function values at the critical points and boundaries of the domain. Note that since \(f(x)\) is increasing in \([-1,2]\) and since the function is not differentiable at \(x=2\), \(x=2\) is the only critical point to be considered. Thus, we should compare the values of the function at \(x=-1\), \(x=2\), and \(x=3\): \(f(-1) = 3(-1)^2 + 12(-1) - 1 = -10\) \(f(2) = 37\) \(f(3) = 37 - 3 = 34\) Since \(f(2)\) is the maximum value among the three, \(f(x)\) has the maximum value at \(x=2\). So, option (D) is true. In conclusion, options (A), (C), and (D) are true, while option (B) is false.

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