Answer :
To find the [tex]$x$[/tex]-coordinates of all relative maxima of the function [tex]\( f(x) = \frac{3}{5} x^5 + 12 x^4 + 60 x^3 + 3 \)[/tex], we need to follow these steps:
1. Find the Critical Points:
- First, we need to find the derivative of the function, [tex]\( f'(x) \)[/tex]. The derivative helps us find critical points where the slope of the tangent is zero.
- Set [tex]\( f'(x) = 0 \)[/tex] and solve for [tex]\( x \)[/tex]. The solutions are the critical points. For this function, the critical points are [tex]\( x = -10 \)[/tex], [tex]\( x = -6 \)[/tex], and [tex]\( x = 0 \)[/tex].
2. Determine the Nature of Each Critical Point:
- Calculate the second derivative, [tex]\( f''(x) \)[/tex], to determine the concavity of the function at these critical points.
- Use the second derivative test: If [tex]\( f''(x) < 0 \)[/tex] at a critical point [tex]\( x = c \)[/tex], then [tex]\( c \)[/tex] is a relative maximum.
3. Evaluate the Second Derivative at Each Critical Point:
- Check [tex]\( f''(x) \)[/tex] at each critical point:
- For [tex]\( x = -10 \)[/tex], the second derivative is negative, indicating a relative maximum.
- For [tex]\( x = -6 \)[/tex] and [tex]\( x = 0 \)[/tex], the second derivative is not negative, indicating these are not relative maxima.
Therefore, the [tex]$x$[/tex]-coordinate of the relative maximum is [tex]\( x = -10 \)[/tex].
1. Find the Critical Points:
- First, we need to find the derivative of the function, [tex]\( f'(x) \)[/tex]. The derivative helps us find critical points where the slope of the tangent is zero.
- Set [tex]\( f'(x) = 0 \)[/tex] and solve for [tex]\( x \)[/tex]. The solutions are the critical points. For this function, the critical points are [tex]\( x = -10 \)[/tex], [tex]\( x = -6 \)[/tex], and [tex]\( x = 0 \)[/tex].
2. Determine the Nature of Each Critical Point:
- Calculate the second derivative, [tex]\( f''(x) \)[/tex], to determine the concavity of the function at these critical points.
- Use the second derivative test: If [tex]\( f''(x) < 0 \)[/tex] at a critical point [tex]\( x = c \)[/tex], then [tex]\( c \)[/tex] is a relative maximum.
3. Evaluate the Second Derivative at Each Critical Point:
- Check [tex]\( f''(x) \)[/tex] at each critical point:
- For [tex]\( x = -10 \)[/tex], the second derivative is negative, indicating a relative maximum.
- For [tex]\( x = -6 \)[/tex] and [tex]\( x = 0 \)[/tex], the second derivative is not negative, indicating these are not relative maxima.
Therefore, the [tex]$x$[/tex]-coordinate of the relative maximum is [tex]\( x = -10 \)[/tex].
