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The graph of \(g(x)=0.2x^5-2x^2+4\) has exactly one point in \([1.5, 2.5]\) where the tangent is parallel to \(h(x)=3x-1\). Use the bisection method to approximate the x-coordinate \(x^*\) to two decimal places.
Hints
- Parallel lines have equal slopes.
- Set the first derivative equal to the line's slope.
- Use endpoint signs to decide which half-interval still contains the zero.
- Continue until the interval determines the answer to two decimal places.
Solution
1. Differentiate: \(g'(x)=x^4-4x\). The line \(h\) has slope \(3\), so solve \(F(x)=x^4-4x-3=0\).
2. Since \(F(1.5)=-3.9375<0\) and \(F(2.5)=26.0625>0\), a zero lies in \([1.5, 2.5]\).
3. Repeatedly halve the interval and retain the half whose endpoints have opposite signs. The successive brackets include \([1.5, 2]\), \([1.75, 2]\), and \([1.75, 1.875]\).
4. Continuing the bisection process gives \(x^*\approx1.78436\).
5. Rounded to two decimal places, \(x^*\approx1.78\).
Answer
\(x^*\approx1.78\)
