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Chapter 6: Problem 38

A function \(f(x)\) satisfies the relation \(f(x+y)=f(x)+f(y)+x y(x+y), \forall x, y \in R\). If \(f^{\prime}(0)=-1\), then (a) \(f(x)\) is a polynomial function (b) \(f(x)\) is an exponential function (c) \(f(x)\) is twice differentiable for all \(x \in R\) (d) \(f^{\prime}(3)=8\)

Short Answer

Expert verified
(a) and (c) are correct, while (d) is not true.

Step by step solution

01

Substitute y = 0 into the functional equation

We start by substituting \(y = 0\) into the given functional equation: \(f(x+0) = f(x) + f(0) + x \cdot 0 \cdot x\). This simplifies to \(f(x) = f(x) + f(0)\). Hence, \(f(0) = 0\).
02

Differentiate the functional equation

To find the derivative \(f'(x)\), we differentiate the functional equation with respect to \(x\): \[\frac{d}{dx} [f(x+y) = f(x) + f(y) + xy(x+y)].\]Using product rule and chain rule, we differentiate: \[f'(x+y) \cdot 1 = f'(x) + 0 + (y(x+y) + xy \cdot 1)\ = f'(x) + y(x+2y).\]
03

Evaluate the derivative at a specific point

We found \(f'(x+y) = f'(x) + y(x+2y)\). Substitute \(y = 0\) and evaluate at \(x = 0\) to use \(f'(0) = -1\): \[f'(0+0) = f'(0) = -1.\]This confirms the condition given in the problem description.
04

Differentiate the derived expression again

Rewriting the expression and differentiating with respect to \(x\) again, recognize it leads to a second derivative: \[f''(x+y) = f''(x) + 2y.\]Thus, \(f''(x)\) is constant, suggesting \(f(x)\) is a quadratic polynomial.
05

Check consistency with polynomial form

Assume \(f(x) = ax^2 + bx + c\) based on constant second derivative. Since \(f(0) = 0\), \(c = 0\). \(f'(0) = -1\) implies \(b = -1\). Further substituting in functional equation, we get \(f(x+y)=a(x+y)^2-b(x+y)\ = ax^2+2axy+ay^2-bx-by\).Compare, coordinate terms and solve to confirm \(a=1\), \(b=-1\), showing \(f(x) = x^2 - x\).
06

Calculate derived expression at a point

Using derived function \(f(x) = x^2 - x\), the derivative \(f'(x) = 2x - 1\). Evaluate \(f'(3) = 2(3) - 1 = 6 - 1 = 5\).
07

Address choices using solution verification

Based on steps, validate given choices: - (a) True, as \(f(x)\) is polynomial.- (c) True, since \(f(x) = x^2 - x\) is twice differentiable everywhere. - Given (d) is incorrect as \(f'(3) = 5\), not 8.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polynomial Function
A polynomial function is an expression made up of variables and coefficients, linked together by addition, subtraction, multiplication, and non-negative integer exponents. Understanding polynomial functions- **Example**: A simple polynomial could be expressed as \( f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \) where \( a_n, a_{n-1}, \ldots, a_0 \) are coefficients and \( n \) is a non-negative integer.- **Characteristics**:
  • Polynomials are smooth and continuous functions.
  • They have a finite number of terms.
  • Each term's degree is determined by the exponent of its variable.
- In the given exercise with the function \( f(x) = x^2 - x \), we can see this is a polynomial function of degree 2 with a leading term of \( x^2 \), indicating that it is specifically a quadratic polynomial.
Differentiation
Differentiation is the process of determining the derivative of a function. It tells us the rate at which one quantity changes with respect to another.- **Definition**: The derivative of a function \( f(x) \) is represented as \( f'(x) \) or \( \frac{df}{dx} \).- **Finding a Derivative**:
  • For \( f(x) = x^2 - x \), using differentiation, the derivative is \( f'(x) = \frac{d}{dx}(x^2 - x) = 2x - 1 \).
- **Purpose**: Derivatives have numerous applications; they help find slopes or tangents to curves, and optimizations in calculus problems.In this exercise, understanding differentiation was key to determining whether \( f(x) \) matched the conditions required by the given problem.
Constant Function
A constant function is a particular type of function that gives the same value for any input value.- **Definition**: A function \( f(x) \) is called constant if it can be written as \( f(x) = c \), where \( c \) is a constant.- **Characteristics**:
  • Inevitably, the derivative of a constant function is zero, \( f'(x) = 0 \), reflecting no change.
  • On graphs, constant functions appear as horizontal lines.
- **In Context**: Even though our main focus function here is not constant, understanding how its components work helps emphasize differences in behavior, such as in quadratic equations where constant second derivatives can lead to identifying degreed polynomial shapes.
Quadratic Function
Quadratic functions have a specific structure involving quadratic terms, making them unique within polynomial functions.- **Form**: A typical quadratic function is expressed as \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are coefficients and \( a eq 0 \).- **Graph**: They form a parabolic shape that opens upward if \( a > 0 \) or downward if \( a < 0 \).- **Differentiation and Properties**:
  • The first derivative \( f'(x) = 2ax + b \) indicates how the slope of the tangent changes at any point on the parabola.
  • In the example function \( f(x) = x^2 - x \), \( a = 1 \), showing it opens upward and has distinctive points and tangent behavior.
Understanding how the quadratic component of a polynomial function behaves allows for deeper insight into solutions and derivations needed for evaluations like finding specific function conditions.

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Most popular questions from this chapter

If \(f(x)=0\) for \(x<0\) and \(f(x)\) is differentiable at \(x=0\), then for \(x>0, f(x)\) may be (a) \(x^{2}\) (b) \(x\) (c) \(-x\) (d) \(-x^{3 / 2}\)

Let \(f_{1}(x)\) and \(f_{2}(x)\) be twice differentiable functions where \(F(x)=f_{1}(x)+f_{2}(x)\) and \(G(x)=f_{1}(x)-f_{2}(x), \forall x \in R, f_{1}(0)=2\) and \(f_{2}(0)=1 .\) If \(f_{1}^{\prime}(x)=f_{2}(x)\) and \(f_{2}^{\prime}(x)=f_{1}(x), \forall x \in R\), then the number of solutions of the equation \((F(x))^{2}=\frac{9 x^{4}}{G(x)}\) is .........

\(f\) is a continuous function in \([a, b], g\) is a continuous function in \([b, c]\), A function \(h(x)\) is defined as \(h(x)=\left\\{\begin{array}{l}f(x) \text { for } x \in[a, b) \\ g(x) \text { for } x \in(b, c]\end{array}\right.\), If \(f(b)=g(b)\), then (a) \(h(x)\) has a removable discontinuity at \(x=b\) (b) \(h(x)\) may or may not be continuous in \([a, c]\) (c) \(h\left(b^{-}\right)=g\left(b^{+}\right)\)and \(h\left(b^{+}\right)=f\left(b^{-}\right)\) (d) \(h\left(b^{+}\right)=g\left(b^{-}\right)\)and \(h\left(b^{-}\right)=f\left(b^{+}\right)\)

Let \(g(x)=\left\\{\begin{array}{cl}3 x^{2}-4 \sqrt{x}+1, & \text { for } x<1 \\\ a x+b, & \text { for } x \geq 1\end{array}\right.\). If \(g(x)\) is the continuous and differentiable for all numbers in its domain, then (a) \(a=b=4\) (b) \(a=b=-4\) (c) \(a=4\) and \(b=-4\) (d) \(a=-4\) and \(b=4\)

A function is defined as \(f(x)=\left\\{\begin{array}{c,}e^{x}, x \leq 0 \\\ |x-1|, x>0\end{array}\right.\), then \(f(x)\) is (a) continuous at \(x=0\) (b) continuous at \(x=1\) (c) differentiable at \(x=0\) (d) differentiable at \(x=1\)
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