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Chapter 1: Problem 18

Write down the value of \(f(2)\) (a) \(f(X)\) is a polynomial of degree \(1 .\) (b) \(f(0)=1\) (c) \(f(1)=2\)

Short Answer

Expert verified
The value of \( f(2) \) is 3.

Step by step solution

01

- Define the general form of the polynomial

Since the polynomial is of degree 1, the general form of the polynomial can be written as: \[ f(x) = ax + b \]
02

- Use the given conditions to set up equations

We know that: \[ f(0) = 1 \] Using the general form, this gives us: \[ a(0) + b = 1 \] Therefore, \[ b = 1 \]
03

- Use the second condition to find 'a'

We also know that: \[ f(1) = 2 \] Substituting into the general form, we get: \[ a(1) + b = 2 \]We already know that \( b = 1 \), so: \[ a + 1 = 2 \] Solving for 'a', we get: \[ a = 1 \]
04

- Write down the specific form of the polynomial

Now that we have both 'a' and 'b', the polynomial can be written as: \[ f(x) = x + 1 \]
05

- Compute f(2)

Substitute \( x = 2 \) into the polynomial: \[ f(2) = 2 + 1 = 3 \]

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Key Concepts

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

degree of polynomial
A polynomial's degree is the highest power of the variable in its expression. For a polynomial of degree 1, it means that the highest power of the variable (typically x) is 1. This kind of polynomial is also known as a linear polynomial because its graph forms a straight line. For example, if you have a polynomial like \( f(x) = ax + b \), it's of degree 1 because 'x' has the highest exponent of 1.
system of equations
A system of equations consists of multiple equations that share the same set of variables. Solving a system of equations means finding values for these variables that make all the equations true simultaneously. In our exercise, we used conditions given (specifically, \( f(0) = 1 \) and \( f(1) = 2 \)) to form a system of equations. You then solve this system to find the values of 'a' and 'b' in the general polynomial form.
substitution method
The substitution method is a way to solve a system of equations. First, you solve one equation for one variable, then substitute this expression into the other equation. In the given problem, after determining \( b = 1 \) from the first condition \( f(0) = 1 \), we substituted 'b' into the second condition \( f(1) = 2 \) to find 'a'. This step-by-step use of substitution simplifies finding the solution.

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Most popular questions from this chapter

(a) \(f(3)=-5\). (b) \(f(x) \equiv x^{2}-6 x+4\)

If the roots of \(x^{2}+p x+q=0\) are \(\alpha\) and \(\beta\), where \(\alpha\) and \(\beta\) are nonzero, form the equation whose roots are \(\frac{2}{\alpha}, \frac{2}{\beta}\). \((\mathrm{U}\) of \(\mathrm{L}) \mathrm{p}\)

If \(\frac{x+p}{(x-1)(x-3)} \equiv \frac{q}{x-1}+\frac{2}{x-3}\), the values of \(p\) and \(q\) are: (a) \(p=-2, q=1\) (b) \(p=2, q=1\). (c) \(p=1, q=-2\) (d) \(p=1, q=1\) (e) \(p=1, q=-1\).

Given that \(g(x) \equiv \frac{5-x}{\left(1+x^{2}\right)(1-x)}\), express \(g(x)\) in partial fractions.

If the equation \(2 x^{2}+3 x+1=0\) has roots \(\alpha, \beta\) the equation whose roots are \(\frac{1}{\alpha}, \frac{1}{\beta}\) is: (a) \(3 x^{2}+2 x+1=0\) (b) \(x^{2}+3 x+2=0\) (c) \(2 x^{2}+x+3=0\) (d) \(x^{2}-3 x+2=0\) (e) none of these.
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