Determining the Number of Zeros of a Function Prove that [latex]x^3+4 x+1=0[/latex] has exactly one solution.

Figure 2.50 makes the result seem reasonable, but how can we be sure there are no other zeros outside of the displayed window? Notice that if [latex]f(x)=x^3+4 x+1[/latex],then the Intermediate Value Theorem guarantees one solution, since [latex]f(-1)=-4<0[/latex] and f(0) = 1 > 0. Further, [latex]f^{\prime}(x)=3 x^2+4>0[/latex] for all x. By Theorem 10.2, if f(x) = 0 had two solutions, then f′(x) = 0 would have at least one solution. However, since [latex]f^{\prime}(x) \neq 0[/latex] for all x, it can’t be true that f(x) = 0 has two (or more) solutions. Therefore, f(x) = 0 has exactly one solution.

Question 2.10.2: Determining the Number of Zeros of a Function Prove that x³ ...

Calculus: Early Transcendental Functions

Determining the Number of Zeros of a Function

Prove that $x^3+4 x+1=0$ has exactly one solution.

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Figure 2.50 makes the result seem reasonable, but how can we be sure there are no other zeros outside of the displayed window? Notice that if $f(x)=x^3+4 x+1$ ,then the Intermediate Value Theorem guarantees one solution, since $f(-1)=-4<0$ and f(0) = 1 > 0. Further,

f^{\prime}(x)=3 x^2+4>0

for all x. By Theorem 10.2, if f(x) = 0 had two solutions, then f′(x) = 0 would have at least one solution. However, since $f^{\prime}(x) \neq 0$ for all x, it can’t be true that f(x) = 0 has two (or more) solutions. Therefore, f(x) = 0 has exactly one solution.