Question

use the intermediate value theorem to show that there is a root of the given equation in the specified interval. ( e^x = 6 - 5x, (0, 1) ) the equation ( e^x = 6 - 5x ) is equivalent to the equation ( f(x) = e^x - 6 + 5x = 0 ). ( f(x) ) is continuous on the interval (0, 1), ( f(0) = square ), and ( f(1) = square ). since (square < 0 < square) there is a number ( c ) in (0, 1)) such that ( f(c) = 0 ) by the intermediate value theorem. thus, there is a root of the equation ( e^x = 6 - 5x ), in the interval ((0, 1)).

Explanation:

Step1: Calculate \( f(0) \)

We have the function \( f(x) = e^x - 6 + 5x \). To find \( f(0) \), substitute \( x = 0 \) into the function.
\( f(0)=e^{0}-6 + 5\times0 \)
Since \( e^{0}=1 \) and \( 5\times0 = 0 \), we get \( f(0)=1 - 6+0=- 5 \)

Step2: Calculate \( f(1) \)

To find \( f(1) \), substitute \( x = 1 \) into the function \( f(x)=e^{x}-6 + 5x \)
\( f(1)=e^{1}-6 + 5\times1 \)
\( e^{1}\approx2.718 \), \( 5\times1 = 5 \), so \( f(1)=e - 6 + 5=e - 1\approx2.718-1 = 1.718\) (or we can keep it as \( e - 1 \))

Answer:

\( f(0)=\boxed{-5} \), \( f(1)=\boxed{e - 1} \) (or approximately \( \boxed{1.718} \))