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I'm not sure what level of the Calculus you are, but you can solve this problem either by substituting values into the functions or by taking derivatives.  I'm not sure what your instructor means by "sign charts" because it's not a standard mathematical term, however you can probably translate from the solution below.

In part A, I understand the funciton to by y = 2(sqrt x) -3x. Is that correct? the interval is [1,4]. In this case you can make a table of the value of the function when x is 1, 2, 3, 4. In that range it is always decreasing, so the maximum would be at x=1 and the minimum at x=4. If you know derivatives, then the derivative of the function y is 1/(sqrt(x)) - 3. In the interval [1,4] this derivative is always negative, so the function is decreasing on the interval and the maximum is at x=1 and the minimum is at x=4.

In part B, I understand the function to be cuberoot(x+1) + cuberoot(1-x). Is this correct? The interval is [-1,1].  If you compute the function at -1, 0, and 1 it has a smaller value at -1 than it does at 0 and a smaller values at 1 than it does at 0. And if we plot the function at x=.1 and x=-.1 we find that it's less than the 
value at 0 at -.1 and  also less at .1. The function is symmetric around 0, so the function seems to have a minimum at X=-1 and x=1 and a maximum at x=0.  If you know derivatives, then the derivative of the function is (x+1)^(-2/3) - (1-x)^(-2/3). This derivative function is 0 at the value 0, undefined at x=-1 and undefined at x=1.  It goes to infinity as x approaches -1 and to -infinity when x approaches 1.  So the maximum is at x=0 and the minimum at x=1 and x=-1.
In connection with this problem, I'd like to introduce you to a great graphing site,  Take a look!