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Question: The graph of the hyperbolic tangent function for all real values is: A. Strictly increasing B....

The graph of the hyperbolic tangent function for all real values is:
A. Strictly increasing
B. Strictly decreasing
C. Strictly increasing in the interval [0,)\left[ {0,\infty } \right) and strictly decreasing in the interval (,0]\left( { - \infty ,0} \right]
D. Strictly increasing in the interval (,0]\left( { - \infty ,0} \right] and Strictly decreasing in the interval [0,)\left[ {0,\infty } \right)

Explanation

Solution

We will write the value of tanhx\tanh x as the ratio of hyperbolic sine function to hyperbolic cosine function. We will then draw a corresponding graph of the given function. Then, check the nature of the graph, that is where the graph is increasing or where the graph is decreasing.

Complete step by step solution:
The hyperbolic tangent function is the ratio of hyperbolic sine function to hyperbolic cosine function.
That is, tanhx=sinhxcoshx\tanh x = \dfrac{{\sinh x}}{{\cosh x}}
Now, the value of sinhx=exex2\sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2} and the value of coshx=ex+ex2\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2}
Then, we can write the hyperbolic function of tangent as
tanhx=exex2ex+ex2=exexex+ex\tanh x = \dfrac{{\dfrac{{{e^x} - {e^{ - x}}}}{2}}}{{\dfrac{{{e^x} + {e^{ - x}}}}{2}}} = \dfrac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}
We will plot the graph of tanhx\tanh x to identify the nature of the function.

We can see that as the value of xx increases, then the value of tanhx\tanh x also increases.

Therefore, the graph of the hyperbolic tangent function for all real values is strictly increasing.

Note:
The value of tanhx=exexex+ex\tanh x = \dfrac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}. When xxtends to infinity, tanhx\tanh x tends to 1 such as
limxtanhx=limxexexex+ex =limx(1e2x)(1+e2x) =1limxe2x1+limxe2x =1  \mathop {\lim }\limits_{x \to \infty } \tanh x = \mathop {\lim }\limits_{x \to \infty } \dfrac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}} \\\ = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\left( {1 - {e^{ - 2x}}} \right)}}{{\left( {1 + {e^{ - 2x}}} \right)}} \\\ = \dfrac{{1 - \mathop {\lim }\limits_{x \to \infty } {e^{ - 2x}}}}{{1 + \mathop {\lim }\limits_{x \to \infty } {e^{ - 2x}}}} \\\ = 1 \\\
Similarly, when xx tends to - \infty, the value of tanhx\tanh x tends to 1 - 1. Then, we can say that the value becomes almost the same for higher or lower values of xx.