1- Find the critical numbers of the function. (a) f(x) = x^3 + x^2 - 8x + 3 (b)
ID: 2846068 • Letter: 1
Question
1- Find the critical numbers of the function.
(a) f(x) = x^3 + x^2 - 8x + 3
(b) g(x) = x-(sqr3)x
(c) h(x) = (undefined)
2- Find the absolute maximum and absolute minimum values of the function on the given in-
terval.
(a) f(x) = x^3 - 3x^2 + 1, [-1; 4]
(b) g(x) = x +1/x,[0.1,3]
(c) h(x) = cos x - sin x, [0; pi]
3- Suppose that the function f is
f(x) ={Ix-1I,0<x<2
1, x=0,2
(Explain why the function f has an absolute minimum but no absolute maximum on the closed
interval [0; 2].
4- Verify that the function satises the hypotheses of Rolle's Theorem on the given interval.
Then nd all numbers c that satisfy the conclusion.
(a) f(x) = x^2 + 2x + 2, [-2; 0]
(b) g(x) = x^4 - 2x^2 + 1, [-2; 2]
5- Let f(x) = Ix - 1I
(a) Show that there is no point c such that f'(c) = 0 on the interval [0; 2] even though
f(0) = f(2) = 1.
(b) Explain why this does not contradict Rolle's Theorem.
Explanation / Answer
1.
f'(x) = 3x^2+2x-8= 0
=>
x = -2, 4/3 are critical numbers
(b)
g'(x) = 1- 1/3*(x)^(-2/3) => x= 3^(-1.5) is the critical number
2.
f'(x) = 3x^2 -6x
=>
x =0, 2 are critical points
f(0) = 1, f(2) = -3, f(-1) = -3, f(4)= 17
=> global max = 17, global min = -3
(b)
g'(x) =1-1/x^2,
x = 1 is critical point
g(1) = 2, g(0.1) = 10.1, g(3) = 3.33
=>global minimum = 2, global max = 3.33
(c)
h'(x) = -sinx-cosx
=>
x = 3pi/4 is critical point
h(0) = 1, h(3pi/4) = -2^0.5, h(pi) = -1
global maximum = 1, global minimum = -2^0.5
4.
(a)
f(x) = x^2 + 2x +2
=>
f'(x) = 2x+2
=>
f is continuous on [-2,0] and differentiable on (-2,0). =>satisfies the hypothesis of Rolle's theorem
(b)
g(x) = x^4-2x^2 +1
=>
g'(x) = 4x^3 -4x
=>
g is continuos on [-2,2] and differentiable on (-2,2) => satisfies the hypothesis of Rolle's theorem
5.
f(x) = 1-x, for 0<= x <= 1
= x-1, for 1<= x<= 2
f'(x) = -1, for 0<= x <=1
= 1, for 1<= x < = 2
=> f'(c) = 0 has no solution
(b)
It doesnt contrqdict Rolle's theorem since f is not differentiable at x = 1
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