Question

use the intermediate value theorem to show that there is a root of the given equation in the specified interval. ( x^4 + x - 5 = 0, (1, 2) ) ( f(x) = x^4 + x - 5 ) is (\text{--select--}) on the closed interval (1, 2), ( f(1) = square ), and ( f(2) = square ). since ( -3 < \text{?} < 13 ), there is a number ( c ) in ( (1, 2) ) such that ( f(c) = \text{?} ) by the intermediate value theorem. thus, there is a (\text{--select--}) of the equation ( x^4 + x - 5 = 0 ) in the interval ( (1, 2) ).

Explanation:

Step1: Determine continuity of \( f(x) \)

Polynomial functions are continuous everywhere, so \( f(x)=x^4 + x - 5 \) is continuous on \([1,2]\).

Step2: Calculate \( f(1) \)

Substitute \( x = 1 \) into \( f(x) \):
\( f(1)=1^4 + 1 - 5 = 1 + 1 - 5 = -3 \)

Step3: Calculate \( f(2) \)

Substitute \( x = 2 \) into \( f(x) \):
\( f(2)=2^4 + 2 - 5 = 16 + 2 - 5 = 13 \)

Step4: Apply Intermediate Value Theorem

We know \( f(1)= - 3 \), \( f(2)=13 \), and \( 0 \) is between \( - 3 \) and \( 13 \) (i.e., \( -3<0<13 \)). By Intermediate Value Theorem, there exists \( c\in(1,2) \) such that \( f(c) = 0 \). So there is a root of the equation \( x^4 + x - 5 = 0 \) in \( (1,2) \).

Answer:

s (filling the blanks):

• The first "Select" : continuous
• \( f(1)=\boldsymbol{-3} \)
• \( f(2)=\boldsymbol{13} \)
• The "? " (between -3 and 13) : \( 0 \)
• The second "? " ( \( f(c) = \) ) : \( 0 \)
• The last "Select" : root