prove that 1+x < e^x < 1+ xe^x for x>0 using the mean value theorem
To prove the inequalities 1 + x < e^x < 1 + xe^x for x > 0 using the mean value theorem, we can consider the function f(x) = e^x.
First, we will establish the lower bound of the inequality, 1 + x < e^x:
Step 1: Consider the interval [0, x], where 0 < x. By the mean value theorem, there exists a c in the interval (0, x) such that:
f'(c) = (f(x) - f(0))/(x - 0)
f'(c) = (e^x - e^0)/x
f'(c) = (e^x - 1)/x
Step 2: Since c lies in the interval (0, x), we have c > 0 and c < x. Therefore, c > 0 implies e^c > 1.
Step 3: Combining Step 1 and Step 2, we have:
(e^x - 1)/x > 1 ----- (Equation 1)
Step 4: Multiplying both sides of Equation 1 by x, we get:
e^x - 1 > x
Step 5: Adding 1 to both sides of the inequality, we have:
1 + x < e^x
Thus, we have proven that 1 + x < e^x.
Now, let's establish the upper bound of the inequality, e^x < 1 + xe^x:
Step 6: Consider the interval [0, x], where 0 < x. By the mean value theorem, there exists a c in the interval (0, x) such that:
f'(c) = (f(x) - f(0))/(x - 0)
f'(c) = (e^x - e^0)/x
f'(c) = (e^x - 1)/x
Step 7: Since c lies in the interval (0, x), we have c > 0 and c < x. Therefore, c < x implies e^c < e^x.
Step 8: Combining Step 6 and Step 7, we have:
(e^x - 1)/x < e^x
Step 9: Multiplying both sides of the inequality by x, we get:
e^x - 1 < xe^x
Step 10: Adding 1 to both sides of the inequality, we have:
1 + e^x < 1 + xe^x
Thus, we have proven that e^x < 1 + xe^x.
In conclusion, we have proved the inequalities 1 + x < e^x < 1 + xe^x for x > 0 using the mean value theorem.