Mean Value Theorem Example: finding $c$ for $f(x)=\sqrt{4x-3}$ on $[1,3]$

Problem Statement

• We are given the function
  • $f(x)=\sqrt{4x-3}=(4x-3)^{1/2}$
• Interval of interest: $1\le x\le 3$.
• Task: find the value $c$ (with $c\in(1,3)$) that satisfies the Mean Value Theorem (MVT) for the function on this interval.

Mean Value Theorem (MVT) Refresher

• Conditions for applying MVT
  • The function must be continuous on $[a,b]$.
  • The function must be differentiable on $(a,b)$.
• Guarantee: there exists at least one number $c\in(a,b)$ such that
  $f'(c)=\frac{f(b)-f(a)}{b-a}.$
• Geometric meaning
  • The slope of the tangent line at $x=c$ equals the slope of the secant line connecting $(a,f(a))$ and $(b,f(b))$.

Verifying Conditions for $f(x)=\sqrt{4x-3}$

• Domain restriction: $4x-3\ge0 \Rightarrow x\ge3/4$.
• Because 1>3/4, the entire closed interval $[1,3]$ lies inside the function’s domain.
• $f(x)$ is a composition of polynomial and square‐root functions ⇒ continuous on $[1,3]$ and differentiable on $(1,3)$.
• Thus, MVT applies.

Slope of the Secant Line

1. Evaluate the endpoints:
   • $f(3)=\sqrt{4(3)-3}=\sqrt{9}=3.$
   • $f(1)=\sqrt{4(1)-3}=\sqrt{1}=1.$
2. Compute the slope of the secant line connecting $(1,f(1))$ and $(3,f(3))$:
   $\frac{f(3)-f(1)}{3-1}=\frac{3-1}{2}=1.$

Derivative (Slope of Tangent Line)

• Rewrite the function for easier differentiation:
  $f(x)=(4x-3)^{1/2}.$
• Differentiate using the chain rule and power rule:
  $f'(x)=\frac{1}{2}(4x-3)^{-1/2}\cdot4=\frac{2}{\sqrt{4x-3}}.$

Setting Up the MVT Equation

• By MVT, find $c$ such that
  $f'(c)=1.$
• Substitute the derivative:
  $\frac{2}{\sqrt{4c-3}}=1.$

Solving for $c$

1. Multiply both sides by $\sqrt{4c-3}$:
   $2=\sqrt{4c-3}.$
2. Square both sides:
   $4=4c-3.$
3. Add $3$ to each side:
   $7=4c.$
4. Divide by $4$:
   $c=\frac{7}{4}=1.75.$

• Check that $c\in(1,3)$: indeed, $1.75$ lies inside the open interval.

Geometric Interpretation & Visualization

• Points:
  • $A=(1,1)$
  • $B=(3,3)$
• Secant line through $A$ and $B$ has slope $1$.
• There exists at least one point $C=(c,f(c))\approx(1.75,\sqrt{4(1.75)-3}=\sqrt{4})=(1.75,2)$ where
  • The tangent line is parallel to the AB secant.
  • Graphically, the tangent touches the curve with slope $1$ at that $x$ value.

Key Takeaways

• Applying MVT requires verifying continuity and differentiability on the required domains.
• For radical functions, domain restrictions must be respected.
• The algebraic process:
  1. Find the secant slope.
  2. Compute the derivative.
  3. Equate derivative to secant slope and solve.
• The solution $c=7/4$ is unique here because the derivative is monotone over $[1,3]$ (denominator strictly increases).