Graphing functions Use the guidelines of this section to make ...

Question

Graphing functions Use the guidelines of this section to make a complete graph of f.

f(x) = e⁻ˣ²/₂

Accepted Answer

Hello. In this video, we are going to be graphing the given function. F the given function is F of X is equal to 3E raised to the power of negative X 2 divided by 3. In order to graph this function, let's first discuss what the domain of the function is going to be. Notice that we are given an exponential function. For an exponential function, there is no value of X that's going to cause the domain to be undefined. So the domain is going to be all real numbers from negative affinity to positive infinity. Next, let's go ahead and determine the X intercepts. So, in order to determine the X intercepts, we want to find the values of X that cause the function to be zero. The only thing about this function, though, is that for any value of X that we plug into an exponential function, there is no value of X that's going to cause an exponential function to be zero. So, with that being said, there are no X intercepts. But what about Y intercepts? In order to determine the Y intercepts, we want to plug in the value of 0 into the function. F of 0 is going to be 3E, raise of the power of 0 divided by 3. This is going to simplify to be 3E erase the power of zero. E reads the power of 0 is going to simplify to be 1, and 3 multiplied by 1 is going to be 3. So what this tells us is that we have a Y intercept located at the point 0.3. From the four options, let's go ahead and choose which one of these functions have a Y intercept at 0.3. If we take a look at the upper left function, we have a white intercept at -3. So we know that this first graph is not going to be our solution. For the second function, we do have a white intercept at 3, so that's a potential solution. For the third function on the left, notice that the Y intercept is at -3, so this is not going to be a solution to this function. And for the last graph, we do have a white intercept at 3. So this is also a potential solution. Now what we want to go ahead and do is you want to determine the intervals of increasing and decreasing. So, in order to do that, we are going to take the derivative of the given function. Now, again, the given function is Y is equal to 3E, raise the power of negative X 2 divided by 3. In order to take this derivative, we are going to use the exponential rule for the derivative. F of X is going to be 3 multiplied by the derivative of Ease the power of negative X2 divided by 3. In order to take this derivative, we will first take the original exponential value. Ease the power of negative X2 divided by 3. Then we are going to multiply this by the natural logarithm of the base of the function, which in this case is E, and this will be multiplied by the derivative of negative X2 divided by 3. We will now go ahead and simplify this derivative. Ellen of E is going to simplify to be one. And the derivative of negative X 2 divided by 3 is going to simplify to be negative 2 divided by 3 X. So what this means is that the function F of X is going to simplify to be negative to X multiplied by E, raise the power of negative X 2 divided by 3. Now what we want to go ahead and do is we want to find the points or the critical points of the function. In order to find the critical points, we want to find the values of X that cause the function to be zero. We are going to set each value or each factor of X equal to 0. That is going to give us E to the power of negative X2 divided by 3 equal to 0. And -2 X equal to 0. Now, again, there is no value of X that's going to cause an exponential function to be zero, so we will go ahead and disregard this first equation. But for the second equation, if we divide both sides of the equation by -2, we get X's equal to 0. This means that we have a critical point located at the value of 0. If we plot the value 0 on a number line, what we are going to do is we are going to take testing points to the left and to the right of 0 to plug into the derivative. Potential testing points are going to be -1 and 1. If we plug in -1 into the derivative, F-1 is going to produce a positive value. What this means is that the function is going to be increasing to the left of zero. And if we plug in positive 1 into the function, the first derivative, that is going to produce a negative value, and that means that the function is going to be decreasing to the right of zero. Furthermore, because the graph is switching from increasing to decreasing, at the value X is equal to 0, we have a local maximum. If we take a look at the two functions. This function here depicts the value at X is equal to zero to be a local minimum. So this is not going to be the solution. And that means option B, which has 3 at X is equal to 0. As a local maximum, is going to be the solution to this problem. So this is going to be the graph of the given function. So I hope this video helps you in understanding on how to approach this problem, and I will go ahead and see you all in the next video.

Graphing functions Use the guidelines of this section to make a complete graph of f.

f(x) = e⁻ˣ²/₂

Key Concepts

Exponential Functions

Transformation of Functions

Critical Points and Symmetry

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