The hyperbolic cosine function, denoted

Question

The hyperbolic cosine function, denoted $\cosh\left(x\right)$ , is used to model the shape of a hanging cable (a telephone wire, for example). It is defined as $\cosh\left(x\right)=\frac{e^{x}+e^{-x}}{2}$ .

b. Evaluate $hyperbolic cosine of 0$ . Use symmetry and part (a) to sketch a plausible graph for $y = \cosh (x)$ .

Accepted Answer

Below there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Evaluate sine of H 0 given that the definition of the hyperbolic sine function is sine of H X is equal toe of X minus e of negative X divided by 2, and sketch a plausible graph for Y equals sign of H X. Awesome. So first off, we're asked to solve for two separate answers. First, we're asked to evaluate sine of H 0. Given that the definition of this hyperbolic sine function is sign of H X is equal toe the power of X minus e of negative X all divided by 2. So that's our first answer right now. We know that our first answer we're trying to solve for is we're trying to evaluate the sign of H of 0. Awesome. And then our second answer we're trying to solve for is we're trying to sketch a plausible graph for Y equals sign of H of X. OK. So first off, let us start with solving, so let us start to solve this problem by focusing our attention on evaluating the sign of H of 0. So in order to evaluate, and let's write this in blue, so we're evaluating sign of H of 0. Awesome. So since we're evaluating this expression, we must first substitute X equals 0 into the definition of sign of H of X and then simplify. So let's make a little note of this. So now we're substituting X equals 0 into the definition of sign of H of X. Awesome. So when we do that, we will find that. Sign of H of 0, and I'll put parentheses. So this is sign of H 0 where H of 0 is a function. is equal to the power of 0 minus 0, all divided by 2. Which this is all equal to E 0 minus E 0 divided by 2, which is all equal to 0 divided by 2, which is equal to 0. So, therefore, without a doubt, sign of H of 0 is equal to 0. So that's it, we've solved for our first answer. Hooray, we did it. So now, we could start graphing, or rather sketching a plausible graph for, and let's write this in red, we're trying to graph a plausible graph for Y equals sign of H X. Awesome. So when we do that, we need to note that we need to plot the 0.00 which is the origin point, on the graph as calculated from our sign of H 0 equals 0. We need to note also, as we should recall, that for a large positive X, E to the power of X dominates E to the power of negative X, which will make sine of HX grow exponentially. And for a large negative X value, the E negative X will dominate the E to the power of X value, making sign of HX decrease exponentially. So, we also need to recall and note that the graph will be symmetric about the origin because it is an odd function, which means that it will have reflectional symmetry across the origin. So, as a result, we can draw a smooth curve starting from the left and passing through the origin 0.00 and rising to the right, and this will resemble the shape of an elongated S. So, With that said, let us, so I've gone ahead and plugged in our expression, or rather our graph for sign of H 0 equals 0. So we're using this as our backdrop, and we're trying to do a sketch for Y equals sign of H of X. So I plugged it into our graphing or into my graphing calculator to get the following graph, and as we can see, we have an elongated S shape, and in blue, I'll mark with a dot the origin point, which is at 00. And as you could see, One side of the curve will curve upwards, going up along the Y axis in the positive direction, whereas we have another curve that points downwards going towards the negative end of our Y axis, so it's trending downwards. So as you can see, this sketch is the shape of the graph for Y equals sign of H of X, which I wrote it in red, and our curve is also in red in our graph. So with exponential growth, as we should recall, in the positive direction and a mirror image exponential decay in the negative direction, and it will always be passing through the origin. So like we were saying, so we have exponential growth in the positive direction, so that we have our curve curving upwards. And then we have our mirror image exponential decay in the negative direction which shows the curve tending downwards, and it will always pass through the origin point. And that's it. We've officially solved for this problem. Hooray. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. I

Key Concepts

Hyperbolic Functions

Symmetry in Functions

Graphing Techniques

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