UC知道紧急!!! 数学问题!!!!!
来源：百度知道 编辑：UC知道 时间：2024/06/15 22:34:38
1.Find the extreme values of f(x)=2secx-tanx on the interval
[-3.14/4, 3.14/4] and determine where those values occur.
2. Find the extreme values for f(x)=x^4/3 -3x^1/3 on the interval [-1 ,8] and determine where those values occur.
3.Find the extreme values for f(x)={x^2, x<0
f(x)={x^3,x larger than or equal to 0
and dtermine where those values occour.
4.prove that (2,0) is the colsest point on the curve x^2+ y^2 = 4 to ( 4,0)
5.Express the number 25 as a sum of two nonnegative terms whose
product is as large as possible.
6. Find the dimensions of the rectangle of greatest area that can be inscribed in a circle of radius a.
7. A field is to be constructed with 4000 feet of fence as shown. For what value of x will the area be a maximum?
Please show the whose work of each question. THANK YOU~
[-3.14/4, 3.14/4] and determine where those values occur.
2. Find the extreme values for f(x)=x^4/3 -3x^1/3 on the interval [-1 ,8] and determine where those values occur.
3.Find the extreme values for f(x)={x^2, x<0
f(x)={x^3,x larger than or equal to 0
and dtermine where those values occour.
4.prove that (2,0) is the colsest point on the curve x^2+ y^2 = 4 to ( 4,0)
5.Express the number 25 as a sum of two nonnegative terms whose
product is as large as possible.
6. Find the dimensions of the rectangle of greatest area that can be inscribed in a circle of radius a.
7. A field is to be constructed with 4000 feet of fence as shown. For what value of x will the area be a maximum?
Please show the whose work of each question. THANK YOU~
your question is too much to answer here, if
you can understand Chinese or Janpanese, you may "HI" to me and I will answer you in another lanuage soon.
1.we can ealisyfind that df(x)/dx=2secxtanx- （secx）^2，to find extreme values，df(x)/dx=2secxtanx- （secx）^2must be 0，2tanx＝secx
sinx＝1/2，so x＝PI/6，but，wo need to calculate that is extreme，
f（PI/6）＝（3）^(1/2) and(PI/6)on the interval
[-3.14/4, 3.14/4],that is true
2.As last question ，we can find df(x)/dx=4/3 x^(1/3)-x^(-1/3)，we
can't let it be zero，we get that df(x)/dx>0 all the time,so we can tell that f is incresing on the interval [-1 ,8], f will get the maxmium when x=8 and minium when x=-1