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Chapter 3: Problem 87

Show that the function \(f(x)=x^{3}+x+1\) has no relative extrema on \((-\infty, \infty)\).

Short Answer

Expert verified
The function \(f(x) = x^3 + x + 1\) has no relative extrema on \((-\infty, \infty)\) because its first derivative, \(f'(x) = 3x^2 + 1\), is always positive and defined for all real values of x, yielding no critical points.

Step by step solution

01

Find the first derivative of the function

To find the critical points of the function, we need to find its first derivative. The function is \(f(x) = x^3 + x + 1\). Apply the power rule: \(f'(x) = 3x^2 + 1\)
02

Determine if the derivative has any points where it is equal to zero or undefined

We now need to analyze the first derivative that we found in the previous step to find any critical points. \(f'(x) = 3x^2 + 1\) Since \(f'(x)\) is a quadratic function that does not contain any fraction or radical, it is defined for all values of x. Therefore, there are no points where \(f'(x)\) is undefined. Next, we need to determine if \(f'(x)\) has any points where it is equal to zero. To do this, set \(f'(x)\) equal to zero and solve for x: \(3x^2 + 1 = 0\) However, in this case, there are no real solutions for x because for any real value of x, \(3x^2\) will always be non-negative, so \(3x^2 + 1\) will always be greater than 0. Since we cannot find any real values of x that make \(f'(x) = 0\), there are no critical points for this function.
03

Conclusion

Since we could not find any critical points of the function \(f(x) = x^3 + x + 1\), we can conclude that it has no relative extrema on \((-\infty, \infty)\).

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Most popular questions from this chapter

Use Newton's method to obtain an approximation of the root of \(\cos ^{-1} x-x=0\) accurate to three decimal places.

Use Newton's method to approximate the indicated zero of the function. Continue with the iteration until two successive approximations differ by less than \(0.0001\). The zero of \(f(x)=x^{3}+x-4\) between \(x=0\) and \(x=2\). Take \(x_{0}=1\).

Use Newton's method to approximate the indicated zero of the function. Continue with the iteration until two successive approximations differ by less than \(0.0001\). The zero of \(f(x)=x^{5}+x-1\) between \(x=0\) and \(x=1\). Take \(x_{0}=0.5\).

Least Squares Approximation Suppose we are given \(n\) data points $$ P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right), \ldots, P_{n}\left(x_{n}, y_{n}\right) $$ that are scattered about the graph of a straight line with equation \(y=a x\) (see the figure). The error in approximating \(y_{i}\) by the value of the function \(f(x)=a x\) at \(x_{i}\) is $$ \left[y_{i}-f\left(x_{i}\right)\right] \quad 1 \leq i \leq n $$ a. Show that the sum of the squares of the errors in approximating \(y_{i}\) by \(f\left(x_{i}\right)\) for \(1 \leq i \leq n\) is \(g(a)=\left(y_{1}-a x_{1}\right)^{2}+\left(y_{2}-a x_{2}\right)^{2}+\cdots+\left(y_{n}-a x_{n}\right)^{2}\) b. Show that \(g\) is minimized if $$ a=\frac{x_{1} y_{1}+x_{2} y_{2}+\cdots+x_{n} y_{n}}{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}} $$

Use Newton's method to approximate the indicated zero of the function. Continue with the iteration until two successive approximations differ by less than \(0.0001\). The zero of \(f(x)=5 x+\cos x-5\) between \(x=0\) and \(x=1\). Take \(x_{0}=0.5\).
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