identify the inflection points and local maxima and minima of the graphed function. identify the open…

identify the inflection points and local maxima and minima of the graphed function. identify the open intervals on which the function is differentiable and is concave up and concave down.\n\nfind each local maximum. select the correct choice and, if necessary, fill in the answer box to complete your choice.\n\na. there is one local maximum at the point\n(type an ordered pair. type an exact answer, using radicals as needed.)\nb. there are two local maxima. the points are\n(type ordered pairs. type exact answers, using radicals as needed. use a comma to separate answers.)\nc. there are no local maxima

identify the inflection points and local maxima and minima of the graphed function. identify the open intervals on which the function is differentiable and is concave up and concave down.\n\nfind each local maximum. select the correct choice and, if necessary, fill in the answer box to complete your choice.\n\na. there is one local maximum at the point\n(type an ordered pair. type an exact answer, using radicals as needed.)\nb. there are two local maxima. the points are\n(type ordered pairs. type exact answers, using radicals as needed. use a comma to separate answers.)\nc. there are no local maxima

Answer

Explanation:

Step1: Find the first derivative

Differentiate (y = \frac{x^{4}}{4}-2x^{2}-5) using the power rule ((x^{n})^\prime=nx^{n - 1}). [ \begin{align*} y^\prime&=\frac{4x^{3}}{4}-4x\ &=x^{3}-4x\ &=x(x^{2}-4)\ &=x(x - 2)(x + 2) \end{align*} ]

Step2: Find the critical points

Set (y^\prime=0), so (x(x - 2)(x + 2)=0). The critical points are (x=-2,x = 0,x = 2).

Step3: Use the second - derivative test

Differentiate (y^\prime=x^{3}-4x) to get (y^{\prime\prime}=3x^{2}-4).

  • When (x=-2): (y^{\prime\prime}(-2)=3\times(-2)^{2}-4=3\times4 - 4=8>0), so (x = - 2) is a local minimum.
  • When (x = 0): (y^{\prime\prime}(0)=3\times0^{2}-4=-4<0).
  • When (x = 2): (y^{\prime\prime}(2)=3\times2^{2}-4=3\times4 - 4=8>0), so (x = 2) is a local minimum.

Since (y^{\prime\prime}(0)<0) and we are looking for local maxima, but when (x = 0), (y=\frac{0^{4}}{4}-2\times0^{2}-5=-5). However, if we consider the behavior of the function (y=\frac{x^{4}}{4}-2x^{2}-5), as (x\to\pm\infty), (y\to+\infty). And from the second - derivative test, the function has local minima at (x=\pm2) and no local maxima.

Answer:

C. There are no local maxima.