Find the absolute maximum and absolute minimum values of f on the given interval.
f(x) = e−x − e−7x, [0, 1]
f = e^-x - e^-7x
f' = -e^-x + 7e^-7x
= e^-7x (7-e^6x)
f'=0 when e^6x = 7, or x = (ln7)/6
So, evaluate f at the end points and at that x, to get max and min for the interval.
To check, go to
http://rechneronline.de/function-graphs/
and enter
exp(-x) - exp(-7x)
for the function.
Set the x,y intervals to be 0 to 1
To find the absolute maximum and absolute minimum values of f on the interval [0, 1], we can follow these steps:
Step 1: Find the critical points of f(x) by setting its derivative equal to zero and solving for x.
Let's take the derivative of f(x):
f'(x) = -e^(-x) + 7e^(-7x)
Setting f'(x) = 0, we have:
-e^(-x) + 7e^(-7x) = 0
Step 2: Solve the equation -e^(-x) + 7e^(-7x) = 0 for x.
To solve this equation, let's first multiply both sides by e^x to get rid of the negative exponent:
-1 + 7e^(-6x) = 0
Next, add 1 to both sides:
7e^(-6x) = 1
Then divide both sides by 7:
e^(-6x) = 1/7
Take the natural logarithm (ln) of both sides:
-6x = ln(1/7)
Now solve for x by dividing both sides by -6:
x = (1/6) ln(1/7)
Step 3: Evaluate f(x) at the critical points and endpoints of the interval.
We need to evaluate f(x) at the critical point x = (1/6) ln(1/7), as well as the endpoints of the interval, x = 0 and x = 1.
f(0) = e^(-0) - e^(-7*0) = 1 - 1 = 0
f(1) = e^(-1) - e^(-7*1) = 1/e - 1/e^7
Now let's evaluate f(x) at the critical point:
f((1/6) ln(1/7)) = e^(-(1/6) ln(1/7)) - e^(-7(1/6) ln(1/7))
Using the properties of logarithms and exponents, this simplifies to:
f((1/6) ln(1/7)) = (1/7)^(1/6) - (1/7)^((7/6))
Step 4: Determine the absolute maximum and absolute minimum values.
To find the absolute maximum and minimum, we compare the values we calculated.
f(0) = 0
f(1) = 1/e - 1/e^7
f((1/6) ln(1/7)) = (1/7)^(1/6) - (1/7)^((7/6))
We can see that f(0) = 0, f(1) = 1/e - 1/e^7, and f((1/6) ln(1/7)) = (1/7)^(1/6) - (1/7)^((7/6)). We can compare these values to find the absolute maximum and absolute minimum:
The absolute maximum value is the largest among these values.
The absolute minimum value is the smallest among these values.
Please note that the calculations for absolute maximum and minimum involve decimal approximations and cannot be simplified further algebraically.
To find the absolute maximum and minimum values of a function on a closed interval [a, b], we can follow the steps:
1. Find all critical points of the function f(x) on the interval [a, b]. These are the points where the derivative of f(x) is either zero or undefined.
2. Evaluate the function at the critical points and the endpoints of the interval.
3. Compare all the values obtained in step 2 and determine the maximum and minimum values.
Step 1: Find the critical points
To find the critical points, we need to find where the derivative of f(x) is zero or undefined. Let's find the derivative first:
f'(x) = d/dx (e^(-x) - e^(-7x))
Using the chain rule, we have:
f'(x) = -e^(-x) + 7e^(-7x)
Setting f'(x) to zero:
-e^(-x) + 7e^(-7x) = 0
Rearranging the equation:
e^(-x) = 7e^(-7x)
Taking the natural logarithm of both sides:
-x = ln(7) - 7x
6x = ln(7)
x = ln(7)/6
So, the critical point is x = ln(7)/6.
Step 2: Evaluate the function
Now, we evaluate the function f(x) at the critical point and the endpoints of the interval [0, 1]:
f(0) = e^0 - e^(-7*0) = 1 - 1 = 0
f(1) = e^(-1) - e^(-7) (using the function f(x) = e^(-x) - e^(-7x))
f(ln(7)/6) = e^(-ln(7)/6) - e^(-7*ln(7)/6)
Step 3: Compare the values
Comparing the values obtained in Step 2, we find:
f(0) = 0
f(1) ≈ -0.979
f(ln(7)/6) ≈ -0.972
Therefore, the absolute maximum value is 0, and the absolute minimum value is approximately -0.979.