Mathematics

Mathematics

Problem

A function gg satisfies g(1)=−5g(1)=-5 and g(4)=7g(4)=7. Which statement best justifies that the equation g(x)=2g(x)=2 has at least one real solution in the interval [1,4][1,4]? a) Because gg is differentiable on (1,4)(1,4) and g(4)−g(1)≠0g(4)-g(1)\neq 0, a real solution must exist by the Mean Value Theorem. b) Because gg is continuous on [1,4][1,4] and 22 lies between g(1)g(1) and g(4)g(4), the Intermediate Value Theorem guarantees at least one real solution of g(x)=2g(x)=2 in [1,4][1,4]. c) Because gg is increasing on [1,4][1,4], the equation g(x)=2g(x)=2 must have exactly one solution in that interval. d) Because gg is continuous on [1,4][1,4] and g(1)g(4)<0g(1)g(4)<0, the Intermediate Value Theorem ensures g(x)=2g(x)=2 has a solution on [1,4][1,4].
Intuitive explanation

The key here is the Intermediate Value Theorem, which says that if a function is continuous, it must hit every yy-value between its endpoints. Since g(1)=−5g(1)=-5 and g(4)=7g(4)=7, and the target value 22 sits right between −5-5 and 77, a continuous function must cross y=2y=2 somewhere in that interval. You just need to verify continuity and confirm that 22 is actually between the two endpoint values.

Intuitive explanation

The key here is the Intermediate Value Theorem, which says that if a function is continuous, it must hit every yy-value between its endpoints. Since g(1)=−5g(1)=-5 and g(4)=7g(4)=7, and the target value 22 sits right between −5-5 and 77, a continuous function must cross y=2y=2 somewhere in that interval. You just need to verify continuity and confirm that 22 is actually between the two endpoint values.

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