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Question Find the numerical value of tanh⁡(0)\tanh(0) and tanh⁡(1)\tanh(1), rounded to five decimal places.

Studdy Solution

STEP 1

Assumptions
1. The hyperbolic tangent function, denoted as tanh⁡(x)\tanh(x), is defined as tanh⁡(x)=sinh⁡(x)cosh⁡(x)\tanh(x) = \frac{\sinh(x)}{\cosh(x)} where sinh⁡(x)\sinh(x) is the hyperbolic sine of xx and cosh⁡(x)\cosh(x) is the hyperbolic cosine of xx.
2. The hyperbolic sine and cosine functions can be expressed in terms of exponential functions: sinh⁡(x)=ex−e−x2\sinh(x) = \frac{e^x - e^{-x}}{2} and cosh⁡(x)=ex+e−x2\cosh(x) = \frac{e^x + e^{-x}}{2}.
3. The value of tanh⁡(0)\tanh(0) and tanh⁡(1)\tanh(1) can be found by plugging in the values of xx into the definition of tanh⁡(x)\tanh(x).
4. The answers should be rounded to five decimal places.

STEP 2

Calculate tanh⁡(0)\tanh(0) using the definition of the hyperbolic tangent function.
tanh⁡(0)=sinh⁡(0)cosh⁡(0)\tanh(0) = \frac{\sinh(0)}{\cosh(0)}

STEP 3

Compute sinh⁡(0)\sinh(0) and cosh⁡(0)\cosh(0) using the exponential expressions for hyperbolic sine and cosine.
sinh⁡(0)=e0−e−02\sinh(0) = \frac{e^0 - e^{-0}}{2} cosh⁡(0)=e0+e−02\cosh(0) = \frac{e^0 + e^{-0}}{2}

STEP 4

Evaluate the exponential functions at x=0x = 0.
e0=1e^0 = 1 e−0=1e^{-0} = 1

STEP 5

Substitute the values from STEP_4 into the expressions for sinh⁡(0)\sinh(0) and cosh⁡(0)\cosh(0).
sinh⁡(0)=1−12=0\sinh(0) = \frac{1 - 1}{2} = 0 cosh⁡(0)=1+12=1\cosh(0) = \frac{1 + 1}{2} = 1

STEP 6

Now, substitute the values of sinh⁡(0)\sinh(0) and cosh⁡(0)\cosh(0) into the definition of tanh⁡(0)\tanh(0).
tanh⁡(0)=01=0\tanh(0) = \frac{0}{1} = 0

STEP 7

The numerical value of tanh⁡(0)\tanh(0), rounded to five decimal places, is:
tanh⁡(0)=0.00000\tanh(0) = 0.00000

STEP 8

Next, calculate tanh⁡(1)\tanh(1) using the definition of the hyperbolic tangent function.
tanh⁡(1)=sinh⁡(1)cosh⁡(1)\tanh(1) = \frac{\sinh(1)}{\cosh(1)}

STEP 9

Compute sinh⁡(1)\sinh(1) and cosh⁡(1)\cosh(1) using the exponential expressions for hyperbolic sine and cosine.
sinh⁡(1)=e1−e−12\sinh(1) = \frac{e^1 - e^{-1}}{2} cosh⁡(1)=e1+e−12\cosh(1) = \frac{e^1 + e^{-1}}{2}

STEP 10

Evaluate the exponential functions at x=1x = 1.
e1=ee^1 = e e−1=1ee^{-1} = \frac{1}{e}

STEP 11

Substitute the values from STEP_10 into the expressions for sinh⁡(1)\sinh(1) and cosh⁡(1)\cosh(1).
sinh⁡(1)=e−1e2\sinh(1) = \frac{e - \frac{1}{e}}{2} cosh⁡(1)=e+1e2\cosh(1) = \frac{e + \frac{1}{e}}{2}

STEP 12

Calculate the values of sinh⁡(1)\sinh(1) and cosh⁡(1)\cosh(1).
sinh⁡(1)=e−1e2=e2−12e\sinh(1) = \frac{e - \frac{1}{e}}{2} = \frac{e^2 - 1}{2e} cosh⁡(1)=e+1e2=e2+12e\cosh(1) = \frac{e + \frac{1}{e}}{2} = \frac{e^2 + 1}{2e}

STEP 13

Now, substitute the values of sinh⁡(1)\sinh(1) and cosh⁡(1)\cosh(1) into the definition of tanh⁡(1)\tanh(1).
tanh⁡(1)=e2−12ee2+12e\tanh(1) = \frac{\frac{e^2 - 1}{2e}}{\frac{e^2 + 1}{2e}}

STEP 14

Simplify the expression for tanh⁡(1)\tanh(1) by canceling out common factors.
tanh⁡(1)=e2−1e2+1\tanh(1) = \frac{e^2 - 1}{e^2 + 1}

STEP 15

Use a calculator to find the numerical value of tanh⁡(1)\tanh(1).
tanh⁡(1)≈e2−1e2+1\tanh(1) \approx \frac{e^2 - 1}{e^2 + 1}

STEP 16

Round the result to five decimal places.
tanh⁡(1)≈0.76159\tanh(1) \approx 0.76159 (This is an approximate value; the exact value will depend on the calculator used.)
The numerical values of each expression are: (a) tanh⁡(0)=0.00000\tanh(0) = 0.00000 (b) tanh⁡(1)≈0.76159\tanh(1) \approx 0.76159

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