For the function f(x)=3cos(2x),what is the first point where it is increasing the fastest?
Also please tell me how you know or derived that!
Thank you!!! Please answer ASAP
the slope is given by the derivative
f(x) is increasing fastest where the slope is steepest. That is, the derivative has a maximum.
Take a look at the graphs of f and f' and you will see that for this problem, f is increasing fastest when x = 3π/4
http://www.wolframalpha.com/input/?i=plot+y%3D3cos2x,+y%3D-6sin2x
I have no idea what a derivative is. Im in precalculus or advamced functions. That isna calculus topic. Is there any other way to explain it? Thank you
i know you guys are busy but please
In that case, just look at the graph. It is clear that cos(x) is steepest where it crosses the x-axis.
Without calculus, I can't think of an analytical test to prove where the slope is steepest. Are there any similar problems in your text?
not at all but it is on the practice exam he provided.
But your answer makes sense thank you!
To find the first point where a function is increasing the fastest, we need to determine the maximum value of its derivative. In this case, let's find the derivative of the function f(x) = 3cos(2x) first.
Step 1: Find the derivative of f(x) with respect to x.
The derivative of f(x) with respect to x can be found using the chain rule and the derivative of the cosine function:
f'(x) = -2 * 3 * sin(2x)
= -6sin(2x)
Step 2: Set f'(x) equal to zero to find the critical points.
To find where the derivative is equal to zero, we set -6sin(2x) = 0:
-6sin(2x) = 0
Step 3: Solve for x.
To solve for x, we solve the equation -6sin(2x) = 0 for x:
sin(2x) = 0
To find the first point where f(x) is increasing the fastest, we need to identify the critical point (x-value) where the derivative changes from negative to positive. The derivative, f'(x) = -6sin(2x), changes its sign from negative to positive when sin(2x) changes from negative to positive.
Since sin(2x) changes sign at intervals of π, we can set up the equation:
2x = π
Solving for x gives us:
x = π/2
Therefore, the first point where f(x) = 3cos(2x) is increasing the fastest is at x = π/2.
Please note that this explanation assumes you are looking for the fastest rate of increase in terms of absolute value. If you are looking for the fastest rate of increase in terms of positive change only, you would need to consider the intervals where the derivative is positive.