Use the Intermediate Value Theorem to choose the intervals over which the function, x^4 - 2x^2 - 1, is guaranteed to have a zero. Select all that apply.

a. [-2,-1]
b. [-1,0]
c. [0,1]
d. [1,2]

Let f(x) be the given function.
Answer is a and d, because f(-2)<0 and f(-1)>0; and f(1)<0 and f(2)>0, so there is a change of sign, signifying that between the limits there is a value of x where f(x) must be 0 (the curve crosses the x axis).

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berno berno · Answer 2 · 2019-10-18T13:21:05+02:00

Answer:

Options a. and d. are correct

Step-by-step explanation:

Intermediate Value Theorem:

Let f be a continuous function on [tex]\left [ a,b \right ][/tex] such that for a number p, [tex]f(a)<p<f(b)[/tex] then there exists q in [tex]\left ( a,b \right )[/tex] such that f(p) = q

Let f(x) = [tex]x^4 - 2x^2 - 1[/tex]

For a. [-2,-1] :

f(-2)=7>0 and f(-1) = -2 < 0 such that f(-1) < 0 < f(-2) then as per the theorem, there exists a number c in (-2,-1) such that f(c) = 0

For d. [1,2]:

f(1)= - 2<0 and f(2) = 7 > 0 such that f(1) < 0 < f(2) then as per the theorem, there exists a number c in (1,2) such that f(c) = 0

So, options a. and d. are correct

Use the Intermediate Value Theorem to choose the intervals over which the function, x^4 - 2x^2 - 1, is guaranteed to have a zero. Select all that apply. a. [-2,-1] b. [-1,0] c. [0,1] d. [1,2]

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Use the Intermediate Value Theorem to choose the intervals over which the function, x^4 - 2x^2 - 1, is guaranteed to have a zero. Select all that apply.

a. [-2,-1]
b. [-1,0]
c. [0,1]
d. [1,2]