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Chapter 2: Problem 53

Suppose $$ f(x)=x^{2}-6 x+11 $$ Find the smallest number \(b\) such that \(f\) is increasing on the interval \([b, \infty)\).

Short Answer

Expert verified
The smallest number \(b\) such that \(f(x)\) is increasing on the interval \([b, \infty)\) is \(b = 3\).

Step by step solution

01

Find the derivative of the function

To find the critical points, we need to find the derivative of the function \(f(x) = x^2 - 6x + 11\) with respect to \(x\). We do that by applying the power rule for differentiation which states that if \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\). \[ f'(x) = \frac{d}{dx} (x^2 - 6x + 11) \] Using the power rule, \[ f'(x) = 2x - 6 \]
02

Find the critical points

To find the critical points, we need to determine when the derivative \(f'(x)\) is equal to 0 or does not exist. Since our derivative is a linear function, it will exist for all values of \(x\). Thus, we only need to find when it is equal to 0. \[ 2x - 6 = 0 \] Solving for \(x\), \[ x = \frac{6}{2} = 3 \] So, the only critical point is \(x = 3\).
03

Determine if \(f(x)\) is increasing or decreasing for the given interval

We now need to determine in which intervals the function is increasing or decreasing. Since our critical point is \(x=3\), we will test a value on each side of this point to determine the function's behavior. Let's test the intervals \((-\infty, 3)\) and \((3, \infty)\). For \((-\infty, 3)\), let's test \(x = 2\): \[ f'(2) = 2(2) - 6 = -2 \] Since the derivative is negative, the function is decreasing in the interval \((-\infty, 3)\). For \((3, \infty)\), let's test \(x = 4\): \[ f'(4) = 2(4) - 6 = 2 \] Since the derivative is positive, the function is increasing in the interval \((3, \infty)\).
04

Find the smallest value of \(b\) for the increasing interval

As we determined that the function is increasing in the interval \((3, \infty)\), the smallest number \(b\) such that \(f(x)\) is increasing on the interval \([b, \infty)\) is: \[ b = 3 \] Thus, \(f(x)\) is increasing on the interval \([3, \infty)\).

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Most popular questions from this chapter

Suppose \(q\) is a polynomial of degree 5 such that \(q(1)=-3\). Define \(p\) by $$ p(x)=x^{6}+q(x) $$ Explain why \(p\) has at least two zeros.

Without doing any calculations or using a calculator, explain why $$ x^{2}+87559743 x-787727821 $$ has no integer zeros. [Hint: If \(x\) is an odd integer, is the expression above even or odd? If \(x\) is an even integer, is the expression above even or odd?]

Suppose \(q\) is a polynomial of degree 4 such that $$ \begin{array}{r} q(0)=-1 . \text { Define } p \text { by } \\ \qquad p(x)=x^{5}+q(x) . \end{array} $$ Explain why \(p\) has a zero on the interval \((0, \infty)\).

Write the indicated expression as \(a\) polynomial. $$ (q(x))^{2} $$

Write the indicated expression as a ratio of polynomials, assuming that $$ r(x)=\frac{3 x+4}{x^{2}+1}, \quad s(x)=\frac{x^{2}+2}{2 x-1}, \quad t(x)=\frac{5}{4 x^{3}+3} $$. $$ (r \circ t)(x) $$
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