find the maximum value of
f(x)= 3cos(x)-2sin(x)
I know you have to take the derivative but after the derivative I don't know how to solve it.
f'(x)=-3sin(x)-2cos(x)=0
I don't know what to do after this please help
so, solve the equation for f'. You have
3sinx = -2cosx
tanx = -2/3
So, that gives you x for maximum value of f(x). Just plug it in and evaluate f(x).
Of course, you need to recall how to take
sin(arctan(x)) and cos(arctan(x))
But that's just trig, right?
Or, note that
f(x) = -√13 sin(x-θ)
where tanθ = 3/2
Its amplitude is clearly √13
To find the maximum value of a function, you need to determine the critical points where the derivative is equal to zero or does not exist and then check their nature (whether they are maximum or minimum points).
In this case, you correctly found the derivative of the function f(x) as f'(x) = -3sin(x) - 2cos(x).
Now let's solve for the critical points by setting f'(x) equal to zero:
-3sin(x) - 2cos(x) = 0.
To simplify this equation, we can divide both sides by -1:
3sin(x) + 2cos(x) = 0.
Now let's rearrange the equation to isolate sin(x):
3sin(x) = -2cos(x).
Divide both sides of the equation by 3 to get:
sin(x) = (-2/3)cos(x).
Now, recall the trigonometric identity sin(x) = opposite/hypotenuse and cos(x) = adjacent/hypotenuse, where x represents an angle. Using these identities, we can write the equation as:
opposite/hypotenuse = (-2/3)(adjacent/hypotenuse).
Cancelling out the hypotenuse on both sides gives us:
opposite = (-2/3)adjacent.
This equation represents the ratio of the lengths of the sides of a triangle. Without loss of generality, we can assume the adjacent side is 3 units long, which means the opposite side will be -2 units (negative because it is in the opposite direction). Thus, we have:
opposite = -2 and adjacent = 3.
Now, let's determine the quadrant in which this angle lies using the signs of the opposite and adjacent sides. Since the opposite side is negative and the adjacent side is positive, we know that the angle lies in the fourth quadrant.
To find the angle in the fourth quadrant, we can use the inverse tangent function (tan^(-1)). However, because the ratio opposite/adjacent for the angle is (-2/3), we need to use the inverse tangent of the ratio, which is atan(-2/3).
Using a calculator, we can find that atan(-2/3) is approximately -0.588. Hence, x = -0.588 is the angle in the fourth quadrant where the derivative is zero.
Now, to determine whether this critical point corresponds to a maximum or minimum, we can use the second derivative test. However, in this case, we can observe that the second derivative is always negative for any value of x. Therefore, the point x = -0.588 corresponds to a maximum point.
Now that we have the critical point, we can substitute it back into the original function f(x) = 3cos(x) - 2sin(x) to find the maximum value:
f(-0.588) = 3cos(-0.588) - 2sin(-0.588).
Evaluating this expression, we find that the maximum value of f(x) is approximately 3.255.
Therefore, the maximum value of f(x) = 3cos(x) - 2sin(x) is approximately 3.255 at x = -0.588.