48. [tex]f(x)=\frac{x^2}{x^2-1}[/tex] on [-4,4]
Answer (1)
Find the first derivative: f ′ ( x ) = ( x 2 − 1 ) 2 − 2 x .
Determine critical points by setting f ′ ( x ) = 0 , which gives x = 0 .
Analyze the sign of f ′ ( x ) to find intervals of increase and decrease.
Evaluate f ( x ) at critical points and endpoints to find local and global extrema: local minimum at x = 0 , global maxima at x = − 4 and x = 4 . Vertical asymptotes are at x = ± 1 .
The function has a local minimum at x = 0 and global maxima at x = − 4 and x = 4 , so the minimum value is f ( 0 ) = 0 and the maximum value is f ( − 4 ) = f ( 4 ) = 15 16 .
\boxed{\text{Local minimum at } x=0, \text{ global maxima at } x=-4 \text{ and } x=4}
Explanation
Problem Analysis We are given the function f ( x ) = x 2 − 1 x 2 and the interval [ − 4 , 4 ] . We need to analyze this function to find critical points, intervals of increase/decrease, concavity, and any local/global extrema.
Finding the First Derivative First, we find the first derivative of f ( x ) . Using the quotient rule, we have
f ′ ( x ) = ( x 2 − 1 ) 2 ( x 2 − 1 ) ( 2 x ) − x 2 ( 2 x ) = ( x 2 − 1 ) 2 2 x 3 − 2 x − 2 x 3 = ( x 2 − 1 ) 2 − 2 x .
Finding Critical Points Next, we find the critical points by setting f ′ ( x ) = 0 . This occurs when − 2 x = 0 , so x = 0 . Also, f ′ ( x ) is undefined when x = ± 1 , but these are vertical asymptotes of the original function, not critical points where the function is defined. Thus, the only critical point is x = 0 .
Intervals of Increase and Decrease Now, we determine the intervals of increase and decrease by analyzing the sign of f ′ ( x ) .
For x < − 1 , 0"> f ′ ( x ) > 0 , so f ( x ) is increasing.
For − 1 < x < 0 , 0"> f ′ ( x ) > 0 , so f ( x ) is increasing.
For 0 < x < 1 , f ′ ( x ) < 0 , so f ( x ) is decreasing.
For 1"> x > 1 , f ′ ( x ) < 0 , so f ( x ) is decreasing.
Thus, f ( x ) is increasing on ( − ∞ , − 1 ) and ( − 1 , 0 ) , and decreasing on ( 0 , 1 ) and ( 1 , ∞ ) .
Finding the Second Derivative Next, we find the second derivative of f ( x ) . Using the quotient rule, we have
f ′′ ( x ) = ( x 2 − 1 ) 4 ( x 2 − 1 ) 2 ( − 2 ) − ( − 2 x ) ( 2 ( x 2 − 1 ) ( 2 x )) = ( x 2 − 1 ) 4 − 2 ( x 2 − 1 ) 2 + 8 x 2 ( x 2 − 1 ) = ( x 2 − 1 ) 3 − 2 ( x 2 − 1 ) + 8 x 2 = ( x 2 − 1 ) 3 6 x 2 + 2 .
Intervals of Concavity Now, we determine the intervals of concavity by analyzing the sign of f ′′ ( x ) . The numerator 6 x 2 + 2 is always positive. Thus, the sign of f ′′ ( x ) depends on the sign of ( x 2 − 1 ) 3 .
For x < − 1 , 0"> f ′′ ( x ) > 0 , so f ( x ) is concave up.
For − 1 < x < 1 , f ′′ ( x ) < 0 , so f ( x ) is concave down.
For 1"> x > 1 , 0"> f ′′ ( x ) > 0 , so f ( x ) is concave up.
Thus, f ( x ) is concave up on ( − ∞ , − 1 ) and ( 1 , ∞ ) , and concave down on ( − 1 , 1 ) .
Inflection Points Since f ′′ ( x ) changes sign at x = − 1 and x = 1 , these are potential inflection points. However, f ( x ) is not defined at these points, so there are no inflection points.
Local and Global Extrema Now, we evaluate f ( x ) at the critical point and endpoints to find local and global extrema.
f ( 0 ) = 0 2 − 1 0 2 = 0
f ( − 4 ) = ( − 4 ) 2 − 1 ( − 4 ) 2 = 15 16
f ( 4 ) = ( 4 ) 2 − 1 ( 4 ) 2 = 15 16
Since f ( 0 ) = 0 is the minimum value and f ( − 4 ) = f ( 4 ) = 15 16 are the maximum values on the interval [ − 4 , 4 ] , we have a local minimum at x = 0 and global maxima at x = − 4 and x = 4 .
Vertical Asymptotes Finally, we identify any vertical asymptotes within the interval [ − 4 , 4 ] . The denominator of f ( x ) is x 2 − 1 , which is zero when x = ± 1 . Thus, there are vertical asymptotes at x = − 1 and x = 1 .
Summary In summary, the function f ( x ) = x 2 − 1 x 2 has a critical point at x = 0 , vertical asymptotes at x = − 1 and x = 1 , is increasing on ( − ∞ , − 1 ) and ( − 1 , 0 ) , decreasing on ( 0 , 1 ) and ( 1 , ∞ ) , concave up on ( − ∞ , − 1 ) and ( 1 , ∞ ) , concave down on ( − 1 , 1 ) , has a local minimum at x = 0 , and global maxima at x = − 4 and x = 4 .
Final Answer The analysis of the function f ( x ) = x 2 − 1 x 2 on the interval [ − 4 , 4 ] reveals key features such as critical points, intervals of increase/decrease, concavity, and extrema. The local minimum is at x = 0 and the global maxima are at x = − 4 and x = 4 .
Examples
Understanding the behavior of functions like f ( x ) = x 2 − 1 x 2 is crucial in many real-world applications. For instance, in physics, this type of function can model the refractive index of a material as a function of frequency. Knowing the critical points and asymptotes helps engineers design optical devices that operate efficiently within specific frequency ranges. Similarly, in economics, such functions can represent cost-benefit ratios, where understanding the extrema and concavity helps in optimizing resource allocation.
