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Prove Existence of $c\\in(0,1)$ for $c^5 + c - 1 = 1/2$ Using Intermediate Value Theorem
Mathematics
University (Calculus 1)
Question Content
(a) Prove there is a value $c\\in(0,1)$ such that $c^5 + c - 1 = \\frac{1}{2}$.
Correct Answer
Such a value $c\\in(0,1)$ exists, proven via the Intermediate Value Theorem.
Detailed Solution Steps
1
Step 1: Rearrange the given equation to form a continuous function. Subtract $\\frac{1}{2}$ from both sides to get $c^5 + c - 1 - \\frac{1}{2}=0$, so define $f(x)=x^5 + x - \\frac{3}{2}$. We need to prove there exists $c\\in(0,1)$ where $f(c)=0$.
2
Step 2: Verify that $f(x)$ is continuous on the interval $[0,1]$. Polynomial functions are continuous everywhere on the real number line, so $f(x)$ is continuous on $[0,1]$.
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Step 3: Evaluate $f(x)$ at the endpoints of the interval: Calculate $f(0)=0^5 + 0 - \\frac{3}{2}= -\\frac{3}{2}<0$, and $f(1)=1^5 + 1 - \\frac{3}{2}=2 - 1.5=0.5>0$.
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Step 4: Apply the Intermediate Value Theorem. Since $f(x)$ is continuous on $[0,1]$, and $f(0)<0<f(1)$, the theorem guarantees there exists at least one $c\\in(0,1)$ such that $f(c)=0$, which means $c^5 + c - 1 = \\frac{1}{2}$.
Knowledge Points Involved
1
Intermediate Value Theorem (IVT)
The theorem states that if a function $f(x)$ is continuous on a closed interval $[a,b]$, and $N$ is any number between $f(a)$ and $f(b)$, then there exists at least one number $c\\in(a,b)$ such that $f(c)=N$. It is used to prove the existence of roots of functions or values that satisfy specific equations on an interval.
2
Continuity of Polynomial Functions
All polynomial functions of the form $f(x)=a_nx^n+...+a_1x+a_0$ are continuous for all real numbers $x$. This is because the sum, product, and constant multiple of continuous functions are continuous, and basic power functions like $x^n$ are continuous.
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Rearranging Equations to Define Functions for Root-Finding
To prove an equation has a solution on an interval, rearrange the equation to set one side equal to 0, then define that side as a function. Proving the function has a root (output of 0) on the interval is equivalent to proving the original equation has a solution there.
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