/*! 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 16 Find all values of \(x\) such th... [FREE SOLUTION] | 91Ó°ÊÓ

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Find all values of \(x\) such that \(f(x)>0\) and all \(x\) such that \(f(x)<0,\) and sketch the graph of \(f\) $$f(x)=x^{5}+1$$

Short Answer

Expert verified
The function is positive for \(x > -1\) and negative for \(x < -1\).

Step by step solution

01

Determine Critical Points

To solve the inequality \(f(x) = x^5 + 1 > 0\), we first find the critical points by setting the equation equal to zero: \(x^5 + 1 = 0\). Solving for \(x\) gives \(x^5 = -1\). The solution to this equation is \(x = -1\) as it is the only real root.
02

Test Intervals around Critical Point

We need to test intervals around the critical point \(x = -1\) to determine where \(f(x) > 0\) and \(f(x) < 0\). Choose test values in the intervals \((-\infty, -1)\), \((-1, \, 0] \), \((0, \, \, \infty)\).
03

Evaluate Test Intervals

Substitute a test value from each interval into \(f(x)\).- For \(x = -2\) (interval \((-\infty, -1)\)): \((-2)^5 + 1 = -32 + 1 = -31\), so \(f(x) < 0\).- For \(x = 0\) (interval \((-1, 0]\)): \(0^5 + 1 = 1\), so \(f(x) > 0\).- For \(x = 1\) (interval \((0, \, \infty)\): \(1^5 + 1 = 2\), so \(f(x) > 0\).
04

Analyze Results

From evaluating the test intervals, we know:- \(f(x) < 0\) when \(x < -1\).- \(f(x) > 0\) when \(x > -1\) including \(0\).Hence, the function is positive for \(x \in (-1, \, \infty)\), and negative for \(x \in (-\infty, 1)\).
05

Sketch the Graph of \(f(x) = x^5 + 1\)

The function \(f(x) = x^5 + 1\) is a polynomial of degree 5, so it resembles the shape of \(x^5\) shifted up by one unit. For large \(|x|\), the function becomes very large, maintaining the same end behavior as \(x^5\): as \(x \rightarrow \infty\), \(f(x) \rightarrow \infty\) and as \(x \rightarrow -\infty\), \(f(x) \rightarrow -\infty\). The critical point \(x=-1\) marks where the function transitions from negative to positive.

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Key Concepts

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

Polynomial Functions
Polynomial functions are mathematical expressions that involve variables raised to whole number powers, combined using addition or subtraction. They are fundamental in algebra and can be linear, quadratic, cubic, quartic, or even higher degrees. A polynomial of degree 'n' looks like this: \[ f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0 \] where \(a_n\) to \(a_0\) are constants and \(a_n eq 0\). - The polynomial function given in the problem is \(f(x) = x^5 + 1\), which is a fifth-degree polynomial because the highest power of the variable \(x\) is 5. - These functions can have multiple roots, and their graphs can have turning points and points of inflection. - High degree polynomials, like the one here, tend to be intricate, displaying significant fluctuations and complex behaviors. Understanding polynomial functions is essential for further analysis like finding the critical points and sketching graphs of these functions.
Critical Points
Critical points of a function are values in the domain where the function's derivative is zero or undefined. They reveal important features of the graph such as turning points or places where direction changes. - For \(f(x) = x^5 + 1\), finding critical points involves setting the equation to zero: \[x^5 + 1 = 0\] which simplifies to \(x = -1\), indicating it's the only real critical point.- Critical points can identify where a function switches from increasing to decreasing or vice versa. They may also indicate local maxima or minima. - In the context of inequalities, critical points are crucial as they mark regions where the function either changes sign or remains consistent. For our function, \(x = -1\) helped determine intervals of positive and negative values. By solving for the roots and evaluating the intervals, we can make judgments about where the function is greater or less than zero.
Graph Sketching
Graph sketching involves creating a rough visual representation of a function, showing its shape, intercepts, asymptotic behavior, and regions where it is positive or negative. It can provide valuable insight into the function's behavior.- For the polynomial \(f(x) = x^5 + 1\), we start by noting its end behavior due to it being a degree 5 polynomial: - As \(x\rightarrow \infty\), \(f(x)\rightarrow \infty\) - As \(x\rightarrow -\infty\), \(f(x)\rightarrow -\infty\)- The point \(x = -1\) is critical as it marks a transition from negative to positive (a turning point), impacting the sketch's slope. The function is positive in the interval \((-1, \infty)\) and negative for \((-\infty, -1)\).- Since its graph resembles \(x^5\) but shifted up one unit, the general curve will cross the y-axis at 1, meaning \( (0, 1) \) is the y-intercept.By piecing together this critical information, one can sketch the general trend of the function: starting from negative infinity, crossing upwards at the critical point, and continuing upwards as \(x\) grows larger. This visual aid helps students both understand and solve real polynomial functions visually.

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Most popular questions from this chapter

Medicare recipients The function \(f\) given by $$ f(x)=-0.000015 z^{3}-0.005 z^{2}+0.75 z+23.5 $$ where \(z=x-1973,\) approximates the total number of Medicare recipients in millions, from \(x=1973\) to \(x=2005 .\) There were \(23,545,363\) Medicare recipients in 1973 and \(42,394,926\) in 2005 (a) Graph \(f,\) and discuss how the number of Medicare recipients has changed over this time period. (b) Create a linear model similar to \(f\) that approximates the number of Medicare recipients. Which model is more realistic?

Use the graph of \(f\) to approximate all values of \(k\) such that \(f(x)\) is divisible by the given linear polynomial. $$f(x)=x^{3}+k^{3} x^{2}+2 k x-2 k^{4} ; \quad x-1.6$$

Automobile sidd marks The speed \(V\) at which an automobile was traveling before the brakes were applied can sometimes be estimated from the length \(L\) of the skid marks. Assume that \(V\) is directly proportional to the square root of \(L\) (a) Express \(V\) as a function of \(L\) by means of a formula that involves a constant of proportionality \(k\) (b) For a certain automobile on a dry surface, \(L=50 \mathrm{ft}\) when \(V=35 \mathrm{mi} / \mathrm{hr} .\) Find the value of \(k\) in part (a). Estimate the initial speed of the automobile in part (b) if the skid marks are 150 feet long.

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Graph \(f,\) and estimate its zeros. $$f(x)=x^{3}-3 x+1$$

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