Explanation Given: The function is, J ( z ) = 1 1 + e − 1.3 ( z − 4 ) − 1 1 + e 5.2 ... ( 1 ) Here, z is the neuron membrane potential, and the net threshold for a neuron to fire is at z = 4 . Approach: Horizontal asymptotes of the function J ( z ) is calculated by considering limit of J ( z ) as z approaches to ∞ . A point where the concavity changes from up to down or down to up is called a point of inflection. Calculation: Consider, lim z → ∞ J ( z ) = lim z → ∞ ( 1 1 + e − 1.3 ( z − 4 ) − 1 1 + e 5.2 ) = lim z → ∞ ( 1 1 + e − 1.3 ( z − 4 ) ) − 1 1 + e 5.2 = 1 1 + lim z → ∞ e − 1.3 ( z − 4 ) − 1 1 + e 5.2 = 1 − 1 1 + e 5.2 Simplify above, lim z → ∞ J ( z ) = 0.995 The line J ( z ) = 0.995 is a horizontal asymptote. Substitute 0 for z in equation ( 1 ) . J ( 0 ) = 1 1 + e − 1.3 ( 0 − 4 ) − 1 1 + e 5.2 = 1 1 + e 5.2 − 1 1 + e 5.2 = 0 Thus, the graph passes through the origin ( 0 , 0 ) . Now, differentiate equation ( 1 ) with respect to z on both sides, J ′ ( z ) = 1 + e − 1.3 ( z − 4 ) ( 0 ) − 1 ( − 1.3 e − 1.3 ( z − 4 ) ) ( 1 + e − 1.3 ( z − 4 ) ) 2 = 1.3 e − 1.3 ( z − 4 ) ( 1 + e − 1.3 ( z − 4 ) ) 2 Clearly denominator is a positive quantity and the numerator is an exponential function that always takes positive values. Therefore, J ′ ( z ) > 0 for all z . So, J ( z ) is an increasing function on ( 0 , ∞ ) . Differentiate equation ( 1 ) again with respect to z . J ″ ( z ) = 1.3 [ ( 1 + e − 1.3 ( z − 4 ) ) 2 ( − 1

2nd Edition

ISBN: 9780321964038

Author: GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher: Pearson Addison Wesley,

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Chapter 5.4, Problem 46E

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The graph of function J(z)=11+e−1.3(z−4)−11+e5.2.

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Chapter 5.4, Problem 45E

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