/*! 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 46 Suppose \(f\) is a differentiabl... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 46

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 \(f(x)\).

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Most popular questions from this chapter

Given \(\mathrm{f}(\mathrm{x})\) is a differentiable function of \(\mathrm{x}\), satisfying \(f(x) . f(y)=f(x)+f(y)+f(x y)-2\) and that \(f(2)=5\). Then \(\mathrm{f}(3)\) is equal to (A) 10 (B) 24 (C) 15 (D) none

Let \(f(x)\) be differentiable at \(x=h\) then $\lim _{x \rightarrow h} \frac{(x+h) f(x)-2 h f(h)}{x-h}$ is equal to (A) \(f(h)+2 h f^{\prime}\) (h) (B) \(2 f(h)+h f^{\prime}\) (C) \(\mathrm{hf}(\mathrm{h})+2 \mathrm{f}^{\prime}\) (h) (D) \(\mathrm{hf}(\mathrm{h})-2 \mathrm{f}^{\prime}\) (h)

Total number of the points where the function $f(x)=\min \\{|x|-1,|x-2|-1 \mid$ is not differentiable (A) 3 points (B) 4 points (C) 5 points (D) None of these

Let \(\mathrm{f}\) be an injective and differentiable function such that f(x). \(f(y)+2=f(x)+f(y)+f(x y)\) for all non negative real \(x\) and \(y\) with \(f^{\prime}(0)=0, f^{\prime}(1)=2 \neq f(0)\), then (A) \(x f^{\prime}(x)-2 f(x)+2=0\) (B) \(x f^{\prime}(x)+2 f(x)-2=0\) (C) \(\times f^{\prime}(x)-f(x)+1=0\) (D) \(2 f(x)=f^{\prime}(x)+2\)

If for a function \(f(x): f(2)=3, f^{\prime}(2)=4\), then $\lim _{x \rightarrow 2}[f(x)]$, where [. ] denotes the greatest integer function, is (A) 2 (B) 3 (C) 4 (D) dne
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