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Found in: Page 224

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# (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 $$n$$, $${e^x} \ge 1 + x + \frac{{{x^2}}}{{2!}} + \ldots \ldots + \frac{{{x^n}}}{{n!}}$$

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

See the step by step solution

## Step 1: Given data

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

## Step 2: 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.

## Step 3: 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 $$x \ge 0$$.

## Step 4: 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 $$x \ge 0$$.

## Step 5: 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 $$x \ge 0$$.