Determine the range of values of [tex]$x$[/tex] for which [tex]$y=\frac{2 x-4}{x^2+5}$[/tex] is a decreasing function.
Answer (1)
Find the derivative of the function y = x 2 + 5 2 x − 4 using the quotient rule: d x d y = ( x 2 + 5 ) 2 − 2 x 2 + 8 x + 10 .
Set the derivative less than zero to find where the function is decreasing: ( x 2 + 5 ) 2 − 2 x 2 + 8 x + 10 < 0 .
Solve the inequality by analyzing the numerator − 2 x 2 + 8 x + 10 < 0 , which simplifies to 0"> ( x − 5 ) ( x + 1 ) > 0 .
Determine the intervals where the inequality holds: x < − 1 or 5"> x > 5 . The function is decreasing on the intervals ( − ∞ , − 1 ) and ( 5 , ∞ ) , so the final answer is 5}"> x < − 1 or x > 5 .
Explanation
Problem Analysis We are given the function y = x 2 + 5 2 x − 4 and we want to find the range of x values for which the function is decreasing. A function is decreasing when its derivative is negative.
Finding the Derivative First, we need to find the derivative of y with respect to x , denoted as d x d y . We will use the quotient rule: If y = v u , then d x d y = v 2 v d x d u − u d x d v . In this case, u = 2 x − 4 and v = x 2 + 5 .
Calculating Derivatives of u and v Now, we calculate d x d u and d x d v .
d x d u = d x d ( 2 x − 4 ) = 2 d x d v = d x d ( x 2 + 5 ) = 2 x
Applying the Quotient Rule Substitute u , v , d x d u , and d x d v into the quotient rule formula to find d x d y .
d x d y = ( x 2 + 5 ) 2 ( x 2 + 5 ) ( 2 ) − ( 2 x − 4 ) ( 2 x )
Simplifying the Derivative Simplify the expression for d x d y .
d x d y = ( x 2 + 5 ) 2 2 x 2 + 10 − ( 4 x 2 − 8 x ) = ( x 2 + 5 ) 2 2 x 2 + 10 − 4 x 2 + 8 x = ( x 2 + 5 ) 2 − 2 x 2 + 8 x + 10
Finding Intervals Where the Function is Decreasing To find where the function is decreasing, we need to find the intervals where d x d y < 0 . This means we need to solve the inequality ( x 2 + 5 ) 2 − 2 x 2 + 8 x + 10 < 0 .
Since ( x 2 + 5 ) 2 is always positive, we only need to consider the numerator: − 2 x 2 + 8 x + 10 < 0 . Dividing by -2, we get 0"> x 2 − 4 x − 5 > 0 .
Factoring and Finding Critical Points Factor the quadratic expression: 0"> ( x − 5 ) ( x + 1 ) > 0 .
Now, we find the critical points by setting ( x − 5 ) ( x + 1 ) = 0 , which gives x = 5 and x = − 1 .
Analyzing Intervals We analyze the intervals determined by the critical points x = − 1 and x = 5 . We test the intervals ( − ∞ , − 1 ) , ( − 1 , 5 ) , and ( 5 , ∞ ) .
For x < − 1 , let x = − 2 . Then 0"> ( − 2 − 5 ) ( − 2 + 1 ) = ( − 7 ) ( − 1 ) = 7 > 0 . So the function is decreasing in this interval.
For − 1 < x < 5 , let x = 0 . Then ( 0 − 5 ) ( 0 + 1 ) = ( − 5 ) ( 1 ) = − 5 < 0 . So the function is increasing in this interval.
For 5"> x > 5 , let x = 6 . Then 0"> ( 6 − 5 ) ( 6 + 1 ) = ( 1 ) ( 7 ) = 7 > 0 . So the function is decreasing in this interval.
Final Answer Therefore, the function is decreasing when x < − 1 or 5"> x > 5 . In interval notation, the range of x values for which the function is decreasing is ( − ∞ , − 1 ) ∪ ( 5 , ∞ ) .
Final Answer The range of values of x for which the function y = x 2 + 5 2 x − 4 is decreasing is x < − 1 or 5"> x > 5 .
Examples
In physics, understanding where a function is decreasing can help analyze the motion of an object. For example, if the function represents the velocity of a car, finding where the function is decreasing tells us when the car is decelerating. This is crucial for designing safe and efficient transportation systems. Similarly, in economics, if the function represents the profit of a company, finding where the function is decreasing helps identify periods of declining profitability, which can inform strategic decisions.
