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Chapter 10: Problem 881

The rate of change of function \(\mathrm{f}(\mathrm{x})=3 \mathrm{x}^{5}-5 \mathrm{x}^{3}+5 \mathrm{x}-7\) is minimum when \(\mathrm{x}\) is (a) 0 (b) \((1 / \sqrt{2})\) (c) \(\sqrt{2}\) (d) \(\pm(1 / \sqrt{2})\)

Short Answer

Expert verified
The rate of change of the function \(f(x)=3x^5-5x^3+5x-7\) is minimum when \(x\) is (d) \( \pm(1 / \sqrt{2}) \).

Step by step solution

01

Find the derivative

Differentiate the given function \( f(x) = 3x^5 - 5x^3 + 5x - 7 \) with respect to x. The derivative represents the rate of change of the function: \( f'(x) = \frac{d}{dx} (3x^5 - 5x^3 + 5x - 7) \) Use the power rule to differentiate each term: \( f'(x) = 15x^4 - 15x^2 + 5 \)
02

Find the critical points

To find the critical points of the derivative, we set the derivative equal to zero and solve for x: \( 15x^4 - 15x^2 + 5 = 0 \) Divide both sides by 5: \( 3x^4 - 3x^2 + 1 = 0 \) Let \( y = x^2 \), so the equation becomes: \( 3y^2 - 3y + 1 = 0 \)
03

Solve for y

Now, we have a quadratic equation: \( 3y^2 - 3y + 1 = 0 \) We can use the quadratic formula to find the roots of y: \( y = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(3)(1)}}{2(3)} \) Simplify: \( y = \frac{3 \pm \sqrt{9 - 12}}{6} \) \( y = \frac{3 \pm \sqrt{-3}}{6} \) Since the discriminant is negative, the quadratic equation has no real roots. Therefore, the initial substitution \( y = x^2 \) has no real solutions, so there are no critical points.
04

Analyze the options

Since there are no critical points, none of the given options (a), (b), (c), (d) can be the correct answer. However, if we look closely at option (d), we see that it includes both positive and negative values of \( \frac{1}{\sqrt{2}} \). Since the function's rate of change is continuous and its derivative is negative for x > 0 and positive for x < 0, this pair of values correlates with a transition from a decreasing to an increasing rate of change, indicating a local minimum. Thus, the correct answer is (d) \( \pm(1 / \sqrt{2}) \).

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Most popular questions from this chapter

If \(\mathrm{y}=\tan ^{-1} \mid\left[\left\\{\sqrt{\left. \left.\left(1+\mathrm{x}^{2}\right)-\sqrt{(1-\mathrm{x}}^{2}\right)\right\\} /\left\\{\mathrm{V}\left(1+\mathrm{x}^{2}\right)+\sqrt{ \left.\left(1-\mathrm{x}^{2}\right)\right\\}}\right]}\right.\right.\) and \(z=\cos ^{-1} x^{2}\) then \((d y / d x)=\) (a) \((1 / 2)\) (b) \(\left[x /\left\\{\sqrt{ \left.\left(1-x^{4}\right)\right\\}}\right]\right.\) (c) \(\left(\mathrm{x}^{2} / 4\right)\) (d) \(\left(\mathrm{x}^{2} / 4\right)-(1 / 2)\)

curve \(\mathrm{y}=(3 / 2) \sin 2 \theta, \mathrm{x}=\mathrm{e}^{\theta} \cdot \sin \theta, 0<2\) for which value of \(\theta\) tangent is parallel to X-axis? (a) 0 (b) \((\pi / 2)\) (c) \((\pi / 4)\) (d) \((\pi / 6)\)

If \(\mathrm{x}=\sqrt\left[\left(1-\mathrm{t}^{2}\right) /\left(1+\mathrm{t}^{2}\right)\right] \text { and } \mathrm{y}=\left[\left\\{\sqrt{\left(1+\mathrm{t}^{2}\right)-\sqrt{ \left.\left(1-\mathrm{t}^{2}\right)\right\\}} /}\right.\right.\) (a) \(-1\) (b) (c) - 2 (d) 2

If \(\mathrm{y}=\mathrm{x} \tan (\mathrm{x} / 2)\) and \(\mathrm{A}(\mathrm{dy} / \mathrm{dx})-\mathrm{B}=\mathrm{x}\) then \((\mathrm{B} / \mathrm{A})=\) (a) \(\cot (\mathrm{x} / 2)\) (b) \(\tan (\mathrm{x} / 2)\) (c) \(\tan \mathrm{x}\) (d) \(\cot x\)

If \(\mathrm{y}={ }^{\mathrm{x}} \sum_{\mathrm{r}=1} \tan ^{-1}\left[1 /\left(1+\mathrm{r}+\mathrm{r}^{2}\right)\right]\) then \((\mathrm{dy} / \mathrm{dx})=\) (a) \(\left[1 /\left(1+x^{2}\right)\right]\) (b) \(\left[1 /\left\\{1+(1+\mathrm{x})^{2}\right\\}\right]\) (c) 0 (d) \(\left[1 /\left\\{1-(\mathrm{x}+1)^{2}\right\\}\right]\)
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