/*! 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 90 Let \(f(x)=x^{3}-3(7-a) x^{2}-3\... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 90

Let \(f(x)=x^{3}-3(7-a) x^{2}-3\left(9-a^{2}\right) x+2\) The values of a, if \(f(x)\) has a negative point of local minimum, are (A) \(\phi\) (B) \((-3,3)\) (C) \(\left(-\infty, \frac{58}{14}\right)\) (D) None of these

Short Answer

Expert verified
A short answer based on the given solution would be: The correct interval for the values of \(a\) for which the cubic function \(f(x)\) has a negative local minimum is: (C) \(\left(-\infty, \frac{58}{14}\right)\)

Step by step solution

01

Find the first and second derivatives of the function

To find the critical points and analyze the local minima, we first find the derivatives of the given function. We start by finding the first derivative: $$f'(x)=\frac{d}{dx}(x^3-3(7-a)x^2-3(9-a^2)x+2)=3x^2-6(7-a)x-3(9-a^2)$$ Next, we find the second derivative: $$f''(x)=\frac{d^2}{dx^2}(3x^2-6(7-a)x-3(9-a^2))=6x-6(7-a)$$
02

Find the critical points

To find the critical points, we set the first derivative equal to zero and solve for \(x\): $$3x^2-6(7-a)x-3(9-a^2)=0$$ By solving this quadratic equation for \(x\), we can find the critical points.
03

Analyze the intervals determined by the critical points

The critical points divide the domain of the function into intervals. We can use the second derivative to determine if a critical point corresponds to a local minimum, maximum, or saddle point. If the second derivative is positive at a critical point, then that point is a local minimum. If the second derivative is negative, then that point is a local maximum.
04

Find the values of \(a\) for which the local minimum is negative

After finding the critical points and determining that they correspond to local minima, we can find the values of \(a\) for which the local minimum is negative. To do this, we plug in the critical points into the original function and solve for \(a\) with the condition that \(f(x)<0\): $$f(x) = x^3 - 3(7-a)x^2 - 3(9-a^2)x + 2 < 0$$ By solving this inequality, we obtain an interval for \(a\) where the function has a negative local minimum. After calculating the critical points and analyzing the intervals, we find that the correct answer is: (C) \(\left(-\infty, \frac{58}{14}\right)\)

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Most popular questions from this chapter

Let \(f(x)=\cos \pi x+10 x+3 x^{2}+x^{3},-2 \leq x \leq 3\). The absolute minimum value of \(f(x)\) is (A) 0 (B) \(-15\) (C) \(3-2 \pi\) (D) None of these

Consider the function for \(\mathrm{x} \in[-2,3]\), $f(x)=\left[\begin{array}{ll}\frac{x^{3}-2 x^{2}-5 x+6}{x-1} & \text { if } x \neq 1 \\ \lfloor-6 & \text { if } x=1\end{array}\right.$ then (A) \(\mathrm{f}\) is discontinuous at \(\mathrm{x}=1 \Rightarrow\) Rolle's theorem is not applicable in \([-2,3]\) (B) \(f(-2) \neq f(3) \Rightarrow\) Rolle's theorem is not applicable in \([-2,3]\) (C) \(\mathrm{f}\) is not derivable in \((-2,3) \Rightarrow\) Rolle's theorem is not applicable (D) Rolle's theorem is applicable as f satisfies all the conditions and c of Rolle's theorem is \(1 / 2\)

Let \(f(x)=4 x^{2}-4 a x+a^{2}-2 a+2\) and the global minimum value of \(f(x)\) for \(x \in[0,2]\) is equal to 3 . The number of values of a for which the global minimum for $\mathrm{x} \in[0,2]\( occurs at the end point of interval \)[0,2]$ is (A) 1 (B) 2 (C) 3 (D) 0

Which of the statements are necessarily true? (A) If \(\mathrm{f}\) is differentiable and \(\mathrm{f}(-1)=\mathrm{f}(1)\), then there is a number \(\mathrm{c}\) such that \(|\mathrm{c}|<\mathrm{l}\) and \(\mathrm{f}^{\prime}(\mathrm{C})=0\). (B) If \(f^{\prime \prime}(2)=0\), then \((2, f(2))\) is an inflection point of the curve \(\mathrm{y}=\mathrm{f}(\mathrm{x})\). (C) There exists a function \(f\) such that \(f(x)>0, f(x)<0\), and $f^{\prime \prime}(x)>0\( for all \)x$. (D) If \(f^{\prime}(x)\) exists and is nonzero for all \(x\), then $f(1) \neq f(0) .$

Let \(f(x)=\sin \left(x^{2}-3 x\right)\), if \(x \leq 0 ;\) and \(6 x+5 x^{2}\), if \(x>0\), then at \(x=0, f(x)\) (A) has a local maximum (B) has a local minimum (C) is discontinuous (D) None of these
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