Show your work/reasoning. Be neat and orderly. Solve on scratch paper first if n

Question

Show your work/reasoning. Be neat and orderly. Solve on scratch paper first if necessary. Use Calculus, not a graphing calculator, to locate any critical numbers and the absolute extrema (give x- and y-coord) of the function f(x) = (x- 2)^2/3 - 1/4 x - 1 on [0, 50]. You need to explicitly state what the critical numbers and absolute extrema are (if any). Let f(x) = arctan(x) on [0, 1]. If the Mean Value Theorem can be applied, find all values c in the interval (0, 1) such that f'(c) = f(1) -f(0)/1 - 0 Given the graph of y = f(x), which is differentiable everywhere and where grid lines are 1/2 unit apart, estimate the following: f'(1/2) (b) All values of x where f'(x) = 0 D_x{f[f(x)]}|_ x = 1/2 The graphs of y = f(x)and y = f'(x) are given. Label the graph as either f(x) or f'(x). State at least two reasons for your labels.

Explanation / Answer

1. Given, f(x)=(x-2)^(2/3)-x/4-1

So, f’(x)=2/3(x-2)^(-1/3)-1/4

The critical points are given by

f’(x)=0

2/3(x-2)^(-1/3)-1/4=0

2/3(x-2)^(-1/3)=1/4

1/(x-2)^(1/3)=6

So, we get

x=566/27

When x=566/27, we get, y=47/54=0.87037

So, the critical point is (566/27, 47/54)

Now, when x=0, y=(-2)^(2/3)-1=.5874

When x=50, y=(50-2)^(2/3)-50/4-1=4 *6^(2/3) - 27/2=-0.292291

So, the extreme points are (0, .5874) and (50, -0.292291)

We see that f(x)=0.87037 is maximum at x=566/27, this is the absolute maximum.

and absolute minimum is f(x)= -0.292291 at x=50

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.