Solve x^7 + x^5 + x^3 + x = 4.66 for the real value of x.

Step-by-step solution

1. 1

   Show there is exactly one real root. Let $f(x) = x^7 + x^5 + x^3 + x$. Its derivative $f'(x) = 7x^6 + 5x^4 + 3x^2 + 1 \ge 1 > 0$ for every real $x$, so $f$ is strictly increasing on all of $\mathbb{R}$. A strictly increasing continuous function takes each value at most once, and since $f \to \pm\infty$, exactly once. No factoring is needed to know the answer is unique.

2. 2

   Bracket the root.

   $f(1) = 1 + 1 + 1 + 1 = 4, \qquad f(1.1) = 1.948717 + 1.610510 + 1.331 + 1.1 = 5.990227$

   Since $4 < 4.66 < 5.990227$, the root lies in $(1,\ 1.1)$.

3. 3

   Narrow the bracket.

   $f(1.03) = 4.511875, \qquad f(1.04) = 4.697449$

   The target $4.66$ sits between these, so the root is in $(1.03,\ 1.04)$ — and closer to $1.04$, since $4.66$ is nearer $4.697$ than $4.512$.

4. 4

   Bisect to full precision. Halving the interval repeatedly (or applying Newton's method with the derivative above) converges to

   $x \approx 1.038016$

5. 5

   Verify and note the precision needed. $f(1.038016) = 4.660006$ to six decimals. By contrast $f(1.038) = 4.659705$ and $f(1.03805) = 4.660645$, so the answer must be carried to about six significant figures — the function changes by roughly $18.8$ per unit of $x$ near the root, so a $10^{-6}$ error in $x$ is a $2\times10^{-5}$ error in $f$.