Identify the coordinates of any local extrema of the function y=e^x - e^(2x).
Let
f(x)=ex - e2x
the domain of f(x) is (-∞,∞).
Thus the extrema of f(x) can be found at point(s) where f'(x)=0.
f'(x)=ex - 2e2x
and f'(x)=0 when
ex = 2e2x
2ex=1
x=ln(1/2) (only root)
Since f"(x)=ex - 4e2x
and
f"(-ln(1/2)) = -1/2
we conclude that a maximum exists at x=ln(1/2) since f" is negative.
Can you take it from here?
To find the coordinates of local extrema of a function, we need to find its critical points first. A critical point is a point on the graph where the derivative of the function is either zero or undefined.
In our case, we have the function y = e^x - e^(2x). To find the critical points, we'll first differentiate the function with respect to x:
dy/dx = d(e^x - e^(2x))/dx
Taking the derivative of each term separately, we get:
dy/dx = d(e^x)/dx - d(e^(2x))/dx
dy/dx = e^x - 2e^(2x)
Setting dy/dx equal to zero and solving for x:
e^x - 2e^(2x) = 0
To solve this equation, we can factor out e^x:
e^x(1 - 2e^x) = 0
This equation gives us two possibilities for critical points:
1) e^x = 0, which is not possible since e^x is always positive.
2) 1 - 2e^x = 0
Solving the second equation for x:
1 - 2e^x = 0
2e^x = 1
e^x = 1/2
Taking the natural logarithm of both sides:
ln(e^x) = ln(1/2)
x = ln(0.5)
Now that we have the x-coordinate of our critical point, we can substitute it back into the original function to find the y-coordinate:
y = e^x - e^(2x)
y = e^(ln(0.5)) - e^(2 * ln(0.5))
y = 0.5 - 0.25
y = 0.25
Therefore, the coordinates of the local extremum are (ln(0.5), 0.25).
Oh, finding extrema sounds "eXciting"! To identify the local extrema of the function y = e^x - e^(2x), we need to find where the derivative is zero. Let's do some math-y stuff.
Taking the derivative of y with respect to x:
dy/dx = e^x - 2e^(2x)
To find where the derivative is zero, let's set it equal to zero and solve:
e^x - 2e^(2x) = 0
Now, this equation might look like an "eXcruciating" task, but luckily, we can do some factoring here. Dividing both sides by e^(2x):
e^(-x) - 2 = 0
Rearranging:
e^(-x) = 2
And, taking the natural logarithm of both sides:
ln(e^(-x)) = ln(2)
Brace yourself for the big reveal:
-x = ln(2)
Now, multiplying both sides by -1:
x = -ln(2)
So, the x-coordinate of the local extrema is -ln(2).
To find the y-coordinate, we can substitute this x-value back into the original equation:
y = e^(-ln(2)) - e^(2(-ln(2)))
Simplifying that expression:
y = 1/2 - 1/4
Drumroll, please:
y = 1/4
Therefore, the coordinates of the local extrema are (-ln(2), 1/4).
Now that we have found the extrema, don't despair, stay "eXuberant"!
To find the local extrema of the function y = e^x - e^(2x), we need to identify the points where the derivative of the function is equal to zero or does not exist. Let's find the derivative first:
dy/dx = d/dx (e^x - e^(2x))
To differentiate e^x, we can use the chain rule, which states that d/dx (e^u) = e^u * (du/dx). In this case, u = x, so du/dx = 1.
Using the chain rule, we differentiate the first term e^x to get e^x * 1 = e^x.
Next, let's apply the chain rule to the second term, e^(2x). Again, using the chain rule, we have d/dx (e^(2x)) = e^(2x) * (d/dx(2x)). The derivative of 2x is 2.
So, for the second term, we have e^(2x) * 2 = 2e^(2x).
Now, let's subtract the two derivatives:
dy/dx = e^x - 2e^(2x).
For local extrema, we need to find the x-values where the derivative dy/dx is equal to zero:
e^x - 2e^(2x) = 0.
To solve this equation, we can factor out e^x:
e^x * (1 - 2e^x) = 0.
This equation is satisfied when either e^x = 0 or 1 - 2e^x = 0.
However, e^x can never be equal to zero, so we only need to solve the second equation:
1 - 2e^x = 0.
Adding 2e^x to both sides:
2e^x = 1.
Dividing both sides by 2:
e^x = 1/2.
Taking the natural logarithm (ln) of both sides:
x = ln(1/2).
Using a calculator, ln(1/2) is approximately -0.6931.
So, the x-coordinate of the local extremum is -0.6931.
To find the corresponding y-coordinate, substitute this value into the original function:
y = e^x - e^(2x).
Using a calculator, we can evaluate the function at x = -0.6931:
y ≈ -0.3491.
Therefore, the coordinates of the local extremum are (-0.6931, -0.3491).