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Let \(f(x)=e^x-2x^2\). Its zeros solve \(e^x=2x^2\).
a) Prove that \(f\) has exactly one zero in \([-1,0]\).
b) Evaluate \(f(1)\), \(f(2)\), and \(f(3)\) to show that the equation has at least two positive solutions.
c) Use derivatives and Rolle’s theorem to show that \(f\) has at most three zeros in total.
Hints
- Use a sign change and monotonicity for part a.
- Use continuity and sign changes for part b.
- Repeatedly apply the consequence of Rolle’s theorem: between two zeros of a function lies a zero of its derivative.
Solution
1. For \(x<0\), \(f'(x)=e^x-4x>0\), so \(f\) is strictly increasing there. Also, \(f(-1)=e^{-1}-2<0\) and \(f(0)=1>0\). By the Intermediate Value Theorem and strict monotonicity, there is exactly one zero in \([-1,0]\).
2. \(f(1)=e-2>0\), \(f(2)=e^2-8<0\), and \(f(3)=e^3-18>0\). Continuity gives one zero in \((1,2)\) and another in \((2,3)\).
3. The derivatives are \(f'(x)=e^x-4x\), \(f''(x)=e^x-4\), and \(f'''(x)=e^x\). Since \(f'''\) has no zeros, Rolle’s theorem implies that \(f''\) has at most one zero, \(f'\) has at most two zeros, and \(f\) has at most three zeros.
Answer
a) Exactly one zero lies in \([-1,0]\).
b) There is at least one zero in each of \((1,2)\) and \((2,3)\).
c) \(f\) has at most three zeros, so these three are all of its real zeros.
