Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Suppose \(f\) is a differentiable function such that \(f(x+y)=f(x)+f(y)+5 x y\) for all \(x, y\) and \(f^{\prime}(0)=3\). The minimum value of \(f(x)\) is (A) \(-1\) (B) \(-9 / 10\) (C) \(-9 / 25\) (D) None

Short Answer

Expert verified
Answer: There is not enough information given to determine the minimum value of \(f(x)\).

Step by step solution

01

Differentiating the given condition with respect to x

To find the minimum value of \(f(x)\), we need to find the first and second derivatives. Differentiating the given condition, we have: \(\frac{\partial}{\partial x}(f(x+y)) = \frac{\partial}{\partial x}(f(x) + f(y) + 5xy)\) Since \(y\) is a constant with respect to \(x\), we have: \(f'(x+y) = f'(x) + 5y\)
02

Substitute to find \(f'(y)\)

Now, we can substitute \(x = 0\) and change \(f'(0) = 3\), to get: \(f'(y) = f'(0) + 5y\) Which gives: \(f'(y) = 3 + 5y\)
03

Find the second derivative

Now, we will find the second derivative by differentiating \(f'(y)\) with respect to \(y\): \(f''(y) = \frac{\partial}{\partial y}(3 + 5y)\) Since the derivative of a constant (\(3\)) is zero, we only need to differentiate \(5y\): \(f''(y) = 5\)
04

Second Derivative Test

Since \(f''(y) = 5\), we can see that the second derivative is positive for all y values. Hence the function is concave up, and we can find the minimum value for \(f(x)\) by taking the critical points of the first derivative: \(f'(y) = 3 + 5y = 0 \Rightarrow y = -\frac{3}{5}\)
05

Finding the minimum value

We have found that at \(y = -\frac{3}{5}\), the critical point exists. Therefore the minimum value of \(f(x)\) occurs at \(y = -\frac{3}{5}\). To find the minimum value, use the condition provided in the problem: \(f(x+y) = f(x) + f(y) + 5xy\) Substitute \(y = -\frac{3}{5}\): \(f\left(x-\frac{3}{5}\right) = f(x) + f\left(-\frac{3}{5}\right) - 3x\) At this point, we know the value of \(f'(0)\), but there is not enough information about \(f(x)\) to compute the exact minimum value. Therefore, the correct answer is (D) None. There is not enough information given to determine the minimum value of \(f(x)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let $\mathrm{f}(\mathrm{x})=\lim _{\mathrm{n} \rightarrow \infty} \frac{\left(\mathrm{x}^{2}+2 \mathrm{x}+3+\sin \pi \mathrm{x}\right)^{\mathrm{n}}-1}{\left(\mathrm{x}^{2}+2 \mathrm{x}+3+\sin \pi \mathrm{x}\right)^{\mathrm{n}}+1}$, then (A) \(\mathrm{f}(\mathrm{x})\) is continuous and differentiable for all \(\mathrm{x} \in \mathrm{R}\). (B) \(f(x)\) is continuous but not differentiable for all \(x \in R\). (C) \(\mathrm{f}(\mathrm{x})\) is discontinuous at infinite number of points. (D) \(\mathrm{f}(\mathrm{x})\) is discontinuous at finite number of points.

If $\mathrm{f}(\mathrm{x}+\mathrm{y}+\mathrm{z})=\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{y}) \cdot \mathrm{f}(\mathrm{z})\( for all \)\mathrm{x}, \mathrm{y}, \mathrm{z}$ and \(f(2)=4, f^{\prime}(0)=3\), then \(f^{\prime}(2)\) equals (A) 12 (B) 9 (C) 16 (D) 6

If \(f(x)=\frac{x}{1+e^{1 / x}}, x \neq 0\) and \(f(0)=0\) then, (A) \(f(x)\) is continuous at \(x=0\) and \(f^{\prime}(x)=1\) (B) \(\mathrm{f}(\mathrm{x})\) is discontinuous at \(\mathrm{x}=0\) (C) \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\) and \(\mathrm{f}^{\prime}(\mathrm{x})\) does not exists (D) \(f(x)\) is continuous at \(x=0\) and \(f^{\prime}(x)=0\)

I.et \(\mathrm{f}(\mathrm{x})\) be a function such that \(\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y})\) and \(f(x)=\sin x g(x)\) for all \(x, y \in R\), If \(g(x)\) is a continuous function such that \(\mathrm{g}(0)=\mathrm{K}\), then \(f^{\prime}(\mathrm{x})\) is equal to (A) \(\mathrm{K}\) (B) \(\mathrm{Kx}\) (C) \(\mathrm{Kg}(\mathrm{x})\) (D) none

Let \(f(x)\) be defined for all \(x \in R\) and the continuous. Let $\mathrm{f}(\mathrm{x}+\mathrm{y})-\mathrm{f}(\mathrm{x}-\mathrm{y})=4 \mathrm{xy} \forall \mathrm{x}, \mathrm{y}=\in \mathrm{R}$ and \(f(0)=0\) then (A) \(\mathrm{f}(\mathrm{x})\) is bounded (B) \(f(x)+f\left(\frac{1}{x}\right)=f\left(x+\frac{1}{x}\right)+2\) (C) \(\mathrm{f}(\mathrm{x})+\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\mathrm{f}\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)+2\) (D) none of these

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free