Find the Local Maxima and Minima f(x)=(2x^2+3)/(4x^2+5)

Problem

ƒ(x)=(2*x2+3)/(4*x2+5)

Solution

1. Find the derivative using the quotient rule d()/d(x)u/v=(vd(u)/d(x)−ud(v)/d(x))/(v2)

ƒ(x)′=((4*x2+5)d(2*x2+3)/d(x)−(2*x2+3)d(4*x2+5)/d(x))/((4*x2+5)2)

2. Differentiate the numerator terms.

ƒ(x)′=((4*x2+5)*(4*x)−(2*x2+3)*(8*x))/((4*x2+5)2)

3. Simplify the numerator by expanding and combining like terms.

ƒ(x)′=(16*x3+20*x−(16*x3+24*x))/((4*x2+5)2)

ƒ(x)′=(−4*x)/((4*x2+5)2)

4. Identify critical points by setting the derivative equal to zero.

−4*x=0

x=0

5. Apply the first derivative test to determine the nature of the critical point.
   For x<0 ƒ(x)′>0 (the function is increasing).
   For x>0 ƒ(x)′<0 (the function is decreasing).

6. Conclude that since the derivative changes from positive to negative at x=0 there is a local maximum at that point.

ƒ(0)=(2*(0)2+3)/(4*(0)2+5)

ƒ(0)=3/5

7. Check for local minima by observing the behavior of the function. Since x=0 is the only critical point and the function decreases as |x| increases toward the horizontal asymptote y=1/2 there are no local minima.

Final Answer

Local Maximum: *(0,3/5), Local Minimum: None

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