Answer to Question #310201 in Calculus for Samantha Nicole Co

Question

A. Identify if the equation f(x)=x⁵-2x⁴+x³-3x²-x+5 has a solution on each given closed interval. Show your proof using the intermediate value theorem.

1.[-2,1]

2.[-1,0]

3.[0,1]

4.[1,2]

5.[2,4]

B. Sketch the graph of f(x)=4-x² and then find the absolute extreme values of the interval [-3,1]

Answer 1

A.

Intermediate value theorem:

If f(x) is continuous on [a,b] and k is strictly between f(a) and f(b), then there exists some c in (a,b) where f(c)=k.

Here

f(x) = x⁵-2x⁴+x³-3x²-x+5

f(x) = x⁵-2x⁴+x³-3x²-x+5 being a polynomial function, is continuous everywhere. So if f(a) < 0 < f(b) or f(b) < 0 < f(a) then there is at least one root in [a,b].

To solve the problem let's first find the value of the function at x = -2, -1, 0, 1, 2 and 4.

f(-2) = (-2)⁵-2(-2)⁴+(-2)³-3(-2)²-(-2)+5

= -77

f(-1) = (-1)⁵-2(-1)⁴+(-1)³-3(-1)²-(-1)+5

= -1

f(0) = (0)⁵-2(0)⁴+(0)³-3(0)²-(0)+5

= 5

f(1) = (1)⁵-2(1)⁴+(1)³-3(1)²-(1)+5

= 1

f(2) = (2)⁵-2(2)⁴+(2)³-3(2)²-(2)+5

= -1

f(4) = (4)⁵-2(4)⁴+(4)³-3(4)²-(4)+5

= 529

(1) For interval [-2, 1]

f(-2) = -77

f(1) = 1

Since f(-2) = -77 < 0 < f(1)=1

there is at least one root in this interval.

(2) For interval [-1,0]

f(-1);= -1

f(0) = 5

Since f(-1) = -1 < 0 < f(0)=5

there is at least one root in this interval.

(3) For interval [0,1]

f(0) = 5 ( Positive)

f(1) = 1 ( Positive)

Since signs are same, nothing can be said in this interval.

(4) For interval [1,2]

f(1) = 1 ( Positive)

f(2) = -1( Negative)

Since f(2)= -1 < 0 < f(1)=1 there is at least one root in this interval.

(5) For interval [2,4]

f(2) = -1 (Negative)

f(4) = 529 (Positive)

Since f(2) = -1 < 0 f(4) = 529 there is a root in this interval.

B. f(x) = 4 - x²

Graph of the function is given below...

f(x) = 4 - x²

f'(x) = -2x

f''(x) = -2 < 0

f'(x) = 0 => -2x = 0 => x = 0

So f(x) has a maximum or minimum at x = 0

Since f''(0) = -2 is negative, f(x) has a maximum value at x = 0

So absolute minimum value in [-3,1] is min{f(-3), f(0), f(1) } = min{ 4-(-3)², 4-(0)², 4-(1)²} = min{-5,4,3} = 3

And absolute maximum value in [-3,1] is max{ 4-(-3)², 4-(0)², 4-(1)²} = min{-5,4,3}= 4