dentify the coordinates of any local and absolute extreme points, inflection points, and intervals of…
dentify the coordinates of any local and absolute extreme points, inflection points, and intervals of concavity. graph the function.\n\n$f(x)=\\ln \\left(30-2 x^{2}\\right)$\n\nidentify any local minima. select the correct choice below and, if necessary, fill in the answer box to complete your choice.\n\na. the local minimum/minima is/are located at (type an ordered pair. type an exact answer. use a comma to separate answers as needed.)\nb. there are no local minima
Answer
Explanation:
Step1: Find the domain
For (y = \ln(30 - 2x^{2})), the argument of the logarithm must be positive. So (30-2x^{2}>0), which simplifies to (x^{2}<15) or (-\sqrt{15}<x<\sqrt{15}).
Step2: Find the first - derivative
Using the chain rule, if (y=\ln(u)) and (u = 30 - 2x^{2}), then (y^\prime=\frac{u^\prime}{u}). (u^\prime=-4x), so (y^\prime=\frac{-4x}{30 - 2x^{2}}=\frac{-2x}{15 - x^{2}}). Set (y^\prime = 0), then (-2x=0) gives (x = 0). We can also use the sign - chart for (y^\prime). Let's test intervals:
- For (x\in(-\sqrt{15},0)), say (x=-1), then (y^\prime=\frac{-2\times(-1)}{15-(-1)^{2}}=\frac{2}{14}>0).
- For (x\in(0,\sqrt{15})), say (x = 1), then (y^\prime=\frac{-2\times1}{15 - 1^{2}}=\frac{-2}{14}<0). Since the function changes from increasing ((y^\prime>0)) to decreasing ((y^\prime<0)) at (x = 0), (x = 0) is a local maximum.
Step3: Analyze for local minima
Since the function has only one critical point (x = 0) (from the first - derivative test) and it is a local maximum, there are no local minima.
Answer:
B. There are no local minima.