The function f(x)= (8x-7)e^2x has one critical number. Find it

A critical point in the domain of x is a value of x at which the function is not differentiable or where the derivative is zero.

See: http://en.wikipedia.org/wiki/Critical_point_%28mathematics%29

For the case in point:
f(x)= (8x-7)e^2x
f'(x) can be found using the product rule and the chain rule as
f'(x):=2(8x-3)*e^(2x)
Setting f'(x)=0 and solving for x, we get e^2x=0 or x=3/8.
Since the range of e^2x excludes 0, it is a solution to be rejected. So the unique point required is x=3/8.

To find the critical number of a function, we need to take the derivative of the function and find the value(s) of x where the derivative is equal to zero or undefined.

Let's start by finding the derivative of the given function, f(x)= (8x-7)e^(2x).

Using the product rule, the derivative of f(x) can be found as follows:

f'(x) = (8x-7)'e^(2x) + (8x-7)(e^(2x))'

To find the derivative of (8x-7), which is a simple linear function, we can apply the power rule:

(8x-7)' = 8

To find the derivative of e^(2x), we can apply the chain rule:

(e^(2x))' = 2e^(2x)

Now we can substitute these values back into the expression for f'(x):

f'(x) = 8e^(2x) + (8x-7)(2e^(2x))

Simplifying further:

f'(x) = 8e^(2x) + 16xe^(2x) - 14e^(2x)

Combining like terms:

f'(x) = (16x - 6)e^(2x)

Now we need to find the critical number(s) by setting the derivative equal to zero and solving for x:

(16x - 6)e^(2x) = 0

Since e^(2x) is always positive (because it is an exponential function), we can disregard it when solving for x. So we solve the equation:

16x - 6 = 0

Adding 6 to both sides:

16x = 6

Dividing both sides by 16:

x = 6/16

Simplifying the fraction:

x = 3/8

Therefore, the critical number of the function f(x) = (8x-7)e^(2x) is x = 3/8.