Question

(a) Show that \({e^x} \ge 1 + x\) for \(x \ge 0\).

(b) Deduce that \({e^x} \ge 1 + x + \frac{1}{2}{x^2}\) for \(x \ge 0\).

(c) Use mathematical induction to prove that for \(x \ge 0\) and any positive integer \(n\),

\({e^x} \ge 1 + x + \frac{{{x^2}}}{{2!}} + \ldots \ldots + \frac{{{x^n}}}{{n!}}\)

Short answer

(a) It is proved that \({f^\prime }(x) = {e^x} - 1\).

(b) The resultant answer is \({e^x} - \left( {1 + x + \frac{1}{2}{x^2}} \right) \ge 0\).

(c) It is proved that for \(x \ge 0\) and any positive integer n, \({e^x} \ge \left( {1 + x + \frac{{{x^2}}}{{2!}} + \ldots \ldots + \frac{{{x^n}}}{{n!}}} \right)\).

Step-by-step solution

Given data

The given function is \({e^x} \ge 1 + x\).

Concept of Differentiation

Differentiation is a method of finding the derivative of a function. Differentiation is a process, where we find the instantaneous rate of change in function based on one of its variables.

Substitute the value

(a)

Obtain the derivative of \(f(x)\).

\(\begin{aligned}{l}{f^\prime }(x) &= \frac{d}{{dx}}\left( {{e^x} - (1 + x)} \right)\\{f^\prime }(x) &= {e^x} - (0 + 1)\\{f^\prime }(x) &= {e^x} - 1\end{aligned}\)

Since \({e^x}\) is an increasing function, \({e^x} \ge 1\) for all \(x \ge 0\),

Substitute 0 in \(f(x)\).

\(\begin{aligned}{l}f(0) &= {e^0} - (1 + 0)\\f(0) &= 1 - 1\\f(0) &= 0\end{aligned}\)

Thus, \(f(x) \ge f(0) = 0\) for all \(x \ge 0\), that is,

\(\begin{aligned}{c}{e^x} - (1 + x) &\ge 0\\{e^x} &\ge 1 + x\end{aligned}\)

Thus, it is proved that \({e^x} \ge 1 + x\) for all \(x \ge 0\).

Simplify the expression

(b)

Obtain the derivative of \(f(x)\).

\(\begin{aligned}{l}{f^\prime }(x) &= \frac{d}{{dx}}\left( {{e^x} - \left( {1 + x + \frac{1}{2}{x^2}} \right)} \right)\\{f^\prime }(x) &= {e^x} - \left( {0 + 1 + \frac{1}{2}(2x)} \right)\\{f^\prime }(x) &= {e^x} - (1 + x)\end{aligned}\)

From part (a)\({e^x} \ge 1 + x\)

Substitute 0 in \(f(x)\).

\(\begin{aligned}{l}f(0) &= {e^0} - \left( {1 + 0 + \frac{1}{2}{{(0)}^2}} \right)\\f(0) &= 1 - 1\\f(0) &= 0\end{aligned}\)

Thus, \(f(x) \ge f(0) = 0\) for all \(x \ge 0\), that is,\({e^x} - \left( {1 + x + \frac{1}{2}{x^2}} \right) \ge 0\)

Thus, it is proved that \({e^x} - \left( {1 + x + \frac{1}{2}{x^2}} \right) \ge 0\) for all \(x \ge 0\).

Simplify the expression

(c)

By induction hypothesis, to show the inequality is true when\(n = 1\).

When\(n = 1,f(x) = {e^x} - (1 + x)\)

In part (a) it already proved that the\({e^x} \ge 1 + x\)for all\(x \ge 0\).

So, the inequality is hold for\(n = 1\).

Assume that the inequality is true for any positive integer\(n\).

So,\({e^x} \ge \left( {1 + x + \frac{{{x^2}}}{{2!}} + \ldots \ldots + \frac{{{x^n}}}{{n!}}} \right)\).

Next prove the inequality is true for positive integer\(n + 1\).

The function is\(f(x) = {e^x} - \left( {1 + x + \frac{{{x^2}}}{{2!}} + \ldots \ldots + \frac{{{x^{n + 1}}}}{{(n + 1)!}}} \right)\).

Obtain the derivative of \(f(x)\).

\(\begin{aligned}{l}{f^\prime }(x) &= \frac{d}{{dx}}\left( {{e^x} - \left( {1 + x + \frac{{{x^2}}}{{2!}} + \ldots \ldots + \frac{{{x^{n + 1}}}}{{(n + 1)!}}} \right)} \right)\\{f^\prime }(x) &= {e^x} - \left( {0 + 1 + x + \frac{{{x^2}}}{{2!}} + \ldots \ldots + \frac{{{x^n}}}{{n!}}} \right)\\{f^\prime }(x) &= {e^x} - \left( {1 + x + \frac{{{x^2}}}{{2!}} + \ldots \ldots + \frac{{{x^n}}}{{n!}}} \right)\end{aligned}\)

As per assumption \({e^x} \ge \left( {1 + x + \frac{{{x^2}}}{{2!}} + \ldots \ldots + \frac{{{x^n}}}{{n!}}} \right)\)

Substitute 0 in \(f(x)\).

\(\begin{aligned}{l}f(0) &= {e^0} - \left( {1 + 0 + \frac{{{0^2}}}{{2!}} + \ldots \ldots + \frac{{{0^{n + 1}}}}{{(n + 1)!}}} \right)\\f(0) &= 1 - 1\\f(0) &= 0\end{aligned}\)

Thus, \(f(x) \ge f(0) = 0\) for all \(x \ge 0\), that is, \({e^x} \ge \left( {1 + x + \frac{{{x^2}}}{{2!}} + \ldots \ldots + \frac{{{x^n}}}{{n!}}} \right)\)

Thus, it is proved that \({e^x} \ge \left( {1 + x + \frac{{{x^2}}}{{2!}} + \ldots \ldots + \frac{{{x^n}}}{{n!}}} \right)\) for all \(x \ge 0\).