consider the equation below

f(x)=e^(7x)+e^(-x)

find the intervals on which f is increasing and decreasing (enter your answers using interval notation)

find the local minimum of f.

find the interval on which f is concave up.

To determine the intervals on which the function is increasing and decreasing, we need to find its derivative and analyze its sign. Let's start by finding the derivative of f(x).

Step 1: Find the derivative of e^(7x).
Differentiating e^(7x) with respect to x gives us 7e^(7x).

Step 2: Find the derivative of e^(-x).
Differentiating e^(-x) with respect to x gives us -e^(-x).

Now we can find the derivative of f(x), f'(x):

f'(x) = 7e^(7x) - e^(-x)

To find the intervals where f(x) is increasing and decreasing, we need to examine the sign of f'(x).

1. Find where f'(x) = 0:

7e^(7x) - e^(-x) = 0

We can simplify this equation by multiplying through by e^x:

7e^(8x) - 1 = 0

Solving for e^(8x):

e^(8x) = 1/7

Taking the natural logarithm (ln) of both sides, we get:

8x = ln(1/7)

x = ln(1/7)/8

Step 2: Analyze the sign of f'(x) in the intervals.

To determine the sign of f'(x) in the intervals, we need to pick values within each interval and evaluate f'(x) at those values. However, since the exponential function is always positive, we can simplify this process:

For x < ln(1/7)/8: The term 7e^(7x) is positive, and the term -e^(-x) is also positive. Therefore, f'(x) > 0.

For x > ln(1/7)/8: Both terms, 7e^(7x) and -e^(-x), are positive. Thus, f'(x) > 0.

Based on the above analysis, f(x) is increasing for all real values of x.

Next, let's find the local minimum of f(x). To do so, we need to find the critical points by setting the derivative f'(x) equal to zero:

7e^(7x) - e^(-x) = 0

Once again, multiplying through by e^x:

7e^(8x) - 1 = 0

Solving for e^(8x):

e^(8x) = 1/7

Taking the natural logarithm (ln) of both sides:

8x = ln(1/7)

x = ln(1/7)/8

So, the critical point is x = ln(1/7)/8.

To find the interval on which f(x) is concave up, we need to analyze the second derivative f''(x). Let's find the second derivative:

f''(x) = \[d^2f(x)/dx^2\]

Taking the derivative of f'(x) with respect to x:

f''(x) = 49e^(7x) + e^(-x)

Since e^(7x) and e^(-x) are always positive, the second derivative f''(x) is positive for all real values of x.

Based on this analysis, f(x) is concave up on the entire real number line.

In summary:

- f(x) is increasing for all real values of x.
- The local minimum of f(x) occurs at x = ln(1/7)/8.
- f(x) is concave up on the entire real number line.

To find the intervals on which f(x) = e^(7x) + e^(-x) is increasing and decreasing, we need to determine where its derivative is positive and negative.

Step 1: Find the derivative of f(x):
f'(x) = 7e^(7x) - e^(-x)

Step 2: Set the derivative equal to zero and solve for x to find the critical points:
7e^(7x) - e^(-x) = 0

Step 3: Solve for x:
7e^(7x) = e^(-x)
e^(8x) = 7
8x = ln(7)
x = ln(7)/8 ≈ 0.201

Step 4: Create a number line by choosing test points in each region defined by the critical points and evaluating the derivative at those test points:
Test point in (-∞, ln(7)/8): x = 0
f'(0) = 7e^(0) - e^(-0) = 7 - 1 = 6
This shows that f'(x) is positive in (-∞, ln(7)/8).

Test point in (ln(7)/8, +∞): x = 1
f'(1) = 7e^(7) - e^(-1)
This evaluation is not necessary for determining the intervals where f'(x) is positive or negative. We can directly tell that f'(x) is negative in this interval since e^(7x) is an increasing function and e^(-x) is a decreasing function.

Step 5: Write out the intervals where f(x) is increasing and decreasing, using interval notation:
f(x) is increasing in (-∞, ln(7)/8)
f(x) is decreasing in (ln(7)/8, +∞)

To find the local minimum of f(x), we need to find the critical points and determine which of them correspond to a minimum.

Step 6: Take the second derivative of f(x):
f''(x) = 49e^(7x) + e^(-x)

Step 7: Substitute the critical points we found earlier into the second derivative:
f''(ln(7)/8) = 49e^(7 * ln(7)/8) + e^(-ln(7)/8)
f''(ln(7)/8) = 49(7^(7/8)) + 7^(-1/8) ≈ 3.242

Since the second derivative at x = ln(7)/8 is positive, this means it is concave up.

Thus, the interval on which f(x) = e^(7x) + e^(-x) is concave up is (ln(7)/8, +∞).

http://www.mathsisfun.com/data/function-grapher.php

f' = 7 e^7x - e^-x
when is that zero?

7 e^7x = e^-x
7 e^8x = 1
e^8x = 1/7
8 x = ln(1/7)
x = -.243 at min