and suppose that g''(x) exists for all x in (a,b). Prove that if there are three values of x in [a,b] for which g(x) = 0 then there is at least one value of x in (a,b) such g''(x) = 0
Can someone help me step by step with this? I am lost.
Can someone help me step by step with this? I am lost.
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The proof requires applying Rolle's Theorem twice: once to show that g'(x) = 0 for two values of x on [a, b] and again to show that g''(x) = 0 between the two values of x such that g'(x) = 0.
Suppose that g(x₁) = g(x₂) = g(x₃) = 0 and x₁, x₂, and x₃ lie on the interval [a, b]. Without a loss of generality, assume that x₁ < x₂ < x₃.
By Rolle's Theorem, g'(x₄) = 0 for some x₄ in (x₁, x₂) and g'(x₅) = 0 for some x₅ in (x₂, x₃). Since g'(x₄) = g'(x₅) = 0, Rolle's Theorem guarantees that g''(x₆) = 0 for some x₆ on (x₄, x₅).
Therefore, since (x₄, x₅) lies on [a, b], there exists at least one value of x (we obtained that one such value is x₆) on (a, b) such that g''(x) = 0, as required.
I hope this helps!
Suppose that g(x₁) = g(x₂) = g(x₃) = 0 and x₁, x₂, and x₃ lie on the interval [a, b]. Without a loss of generality, assume that x₁ < x₂ < x₃.
By Rolle's Theorem, g'(x₄) = 0 for some x₄ in (x₁, x₂) and g'(x₅) = 0 for some x₅ in (x₂, x₃). Since g'(x₄) = g'(x₅) = 0, Rolle's Theorem guarantees that g''(x₆) = 0 for some x₆ on (x₄, x₅).
Therefore, since (x₄, x₅) lies on [a, b], there exists at least one value of x (we obtained that one such value is x₆) on (a, b) such that g''(x) = 0, as required.
I hope this helps!
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Let x_1, x_2, and x_3 satisfy the condition that a <= x_1 < x_2 < x_3 <= b and
g(x_1) = g(x_2) = g(x_3) = 0.
We need to use Rolle's Theorem twice, once on g(x) and then once on g'(x). Note that we are given that g is continuous on [a, b] and g''(x) exists on (a, b).
From Rolle's Theorem on g(x), there exists c in the interval (x_1, x_2) such that g'(c) = 0,
and there exists d in the interval (x_2, x_3) such that g'(d) = 0.
Since differentiability implies continuity and g''(x) exists on (a, b), g'(x) is continuous on (a, b). Note that c and d are in (a, b) and c < d, so g'(x) is continuous on [c, d] and g''(x) exists on (c, d).
From Rolle's Theorem on g'(x), g''(x) = 0 for at least one value of x in the interval (c, d), which is contained in the interval (a, b). This completes the proof.
Lord bless you today!
g(x_1) = g(x_2) = g(x_3) = 0.
We need to use Rolle's Theorem twice, once on g(x) and then once on g'(x). Note that we are given that g is continuous on [a, b] and g''(x) exists on (a, b).
From Rolle's Theorem on g(x), there exists c in the interval (x_1, x_2) such that g'(c) = 0,
and there exists d in the interval (x_2, x_3) such that g'(d) = 0.
Since differentiability implies continuity and g''(x) exists on (a, b), g'(x) is continuous on (a, b). Note that c and d are in (a, b) and c < d, so g'(x) is continuous on [c, d] and g''(x) exists on (c, d).
From Rolle's Theorem on g'(x), g''(x) = 0 for at least one value of x in the interval (c, d), which is contained in the interval (a, b). This completes the proof.
Lord bless you today!
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When g'(x)=0 this means g(x) is at a minimum or maximum. g'(x) indicates which one. Since there are 3 values where g(x)=0, there are at least 2 points where g'(x)=0. (g(x) must have at least 2 point where it changes direction.) Since g''(x)=0 when g'(x) changes direction. Therefore, g''(x) must have 2 points where it is 0.