/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 The global minimum value of $\ma... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 16

The global minimum value of $\mathrm{e}^{\left(2 \mathrm{x}^{2}-2 x+1\right) \sin ^{\prime} \mathrm{x}}$ is (A) \(\mathrm{c}\) (B) \(1 / 3\) (C) 1 (D) 0

Short Answer

Expert verified
Based on the given information and the steps taken, it is impossible to determine which one of the given options leads to the global minimum value of the function. We need more information about the function \(\sin'x\) and \(\sin''x\) to be able to find the global minimum value accurately.

Step by step solution

01

Find the first derivative

To find the critical points, we need to take the derivative of the given function with respect to x. The given function is $$f(x) = e^{(2x^2-2x+1)\sin'x}$$ To find \(f'(x)\), we will use the chain rule: $$f'(x) = \frac{d}{dx}\left(e^{(2x^2-2x+1)\sin'x}\right) = e^{(2x^2-2x+1)\sin'x} \cdot \frac{d}{dx}\left((2x^2-2x+1)\sin'x\right)$$ Now we need to find the derivative of \((2x^2-2x+1)\sin'x\), which requires the product rule, since it's a product of two functions: $$\frac{d}{dx}\left((2x^2-2x+1)\sin'x\right) = (4x-2)\sin'x + (2x^2-2x+1)\cdot(\sin''x)$$ Hence, the first derivative is: $$f'(x) = e^{(2x^2-2x+1)\sin'x} \cdot \left((4x-2)\sin'x + (2x^2-2x+1)\cdot(\sin''x)\right)$$
02

Set the first derivative equal to zero and solve for x

We are looking for the critical points, so we will set the first derivative equal to zero. $$f'(x) = 0$$ Notice that the exponential term \(e^{(2x^2-2x+1)\sin'x}\) is always positive. So, for \(f'(x) = 0\), we only need: $$(4x-2)\sin'x + (2x^2-2x+1)\cdot(\sin''x)= 0$$
03

Analyze the second derivative

Before trying to solve directly for \(x\), let's analyze the second derivative of \(f(x)\). It will help us determine if the function has a minimum value. Find \(f''(x)\): $$f''(x) = \frac{d}{dx}\left(f'(x)\right)$$ It is quite complex to compute \(f''(x)\), and it is not needed to determine if our function has a minimum value. We will now check if the given options satisfy the equation in Step 2, which will help us determine the global minimum value of the function.
04

Test the given options for the minimum value

Now, we will examine each option to see if it satisfies \(f'(x) = 0\). (A) \(f'(c)\): It is impossible to determine if this option satisfies the equation. (B) \(f'\left(\frac{1}{3}\right)\) (C) \(f'(1)\) (D) \(f'(0)\) Without knowing the exact values of \(\sin'x\) and \(\sin''x\), it is impossible to determine which option leads to the global minimum value. The best approach for solving this problem in its current form is to find the function for \(\sin'x\) and \(\sin''x\) and use it to determine which option yields the minimum value. Unfortunately, this isn't possible without more information.

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Most popular questions from this chapter

Which of the given equation has exactly one solution in the indicated interval ? (A) \(x^{3}+2 x-3=0 ;[0,1]\) (B) \(\mathrm{e}^{-x}=\mathrm{x}-1 ;[1,2]\) (C) \(x \ln x=3 ;[2,4]\) (D) \(\sin x=3 x-1 ;[-1,1]\)

The set of critical points of the function $f(x)=x-\log x+\int_{2}^{\pi}(1 / z-2-2 \cos 4 z) d z$ is (A) \(\left\\{\frac{\pi}{6}, \frac{n \pi}{2}+\frac{\pi}{6}\right\\}, n \in N\) (B) \(\\{n \pi\\}, n \in N\) (C) $\left\\{\frac{\pi}{2}, \mathrm{n} \pi+\frac{\pi}{6}\right\\}, \mathrm{n} \in \mathrm{N}$ (D) None of these

In which of the following functions Rolle's theorem is applicable - (A) \(f(x)= \begin{cases}x & , 0 \leq x<1 \\ 0 & , x=1\end{cases}\) (B) $f(x)= \begin{cases}\frac{\sin x}{x},-\pi \leq x<0 \\ 0, & x=0\end{cases}$ (C) \(f(x)=\frac{x^{2}-x-6}{x-1}\) on \([-2,3]\) (D) $f(x)= \begin{cases}\frac{x^{3}-2 x^{2}-5 x+6}{x-1} & \text { if } x \neq 1, x \in[-2,3] \\ -6 & \text { if } x=1\end{cases}$

Let \(g(x)=-\frac{f(-1)}{2} x^{2}(x-1)-f(0)\left(x^{2}-1\right)\) \(+\frac{f(1)}{2} x^{2}(x+1)-f^{\prime}(0) x(x-1)(x+1)\) where \(f\) is a thrice differentiable function. Then the correct statements are (A) there exists \(x \in(-1,0)\) such that \(f^{\prime}(x)=g^{\prime}(x)\) (B) there exists \(x \in(0,1)\) such that $f^{\prime \prime}(x)=g^{\prime \prime}(x)$ (C) there exists \(x \in(-1,1)\) such that $f^{\prime \prime}(x)=g^{\prime \prime \prime}(x)$ (D) there exists \(x \in(-1,1)\) such that \(f^{\prime \prime}(x)=3 f(1)-3 f(-1)\) \(-6 \mathrm{f}^{\prime}(0)\)

The set of value of \(\mathrm{c}\) for which $\sin \\{\ln (\cos \mathrm{x}+\mathrm{c})\\}=1$ has at most one solution in \([0, \pi]\) is (A) \((2 \pi, \infty)\) (B) \(\left(\mathrm{e}^{2 n}, \infty\right)\) (C) \(\left(\frac{\mathrm{e}^{2 \pi}+1}{\mathrm{e}^{2 \pi}-1}, \infty\right)\) (D) null set
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