Problem 43 Combine to show that $$ \mat... [FREE SOLUTION]

91影视

Precalculus A Prelude to Calculus

Sheldon Axler

$Math Studyset 91影视 Explanations$ Math

2 Edition

Chapter 3: Problem 43

Combine to show that $$ \mathbf{1}+t0\) Show that if $t>0$, then $e^{t}<(1+t)^{1+t}$.

Short Answer

Expert verified

To prove $e^t < (1+t)^{1+t}$ for $t>0$, we start with the inequality $1+x0$, we get $(1+xt)^t < (e^{xt})^t$, which simplifies to $(1+xt)^t < e^{x(t^2)}$. Substituting $x=1$, we obtain $(1+t)^t < e^{t^2}$. Finally, by raising both sides to the power of $1/t$, we get $(1+t) < e^t$, which is the desired inequality $e^{t}<(1+t)^{1+t}$ for $t>0$.

Step by step solution

Use the given inequality for non-negative x

For non-negative $x$, we have the inequality $1+x0$.

Manipulate the inequality

Let's first consider the inequality $1+x0$, the function $f(x) = x^t$ is an increasing function. Thus, since $1+xt < e^{xt}$ for non-negative $x$, we can use the increasing property of $f(x)$ to derive the following inequality: $(1+xt)^t < (e^{xt})^t$

Simplify and obtain the desired result

Using properties of exponentiation, we can rewrite the inequality obtained in Step 2 as: $(1+xt)^t < e^{x(t^2)}$ Now, let's substitute $x=1$, which gives us: $(1+t)^t < e^{t^2}$ Now, we want to show that $e^{t}<(1+t)^{1+t}$. To do this, let's raise both sides of our current inequality to the power of $1/t$: $((1+t)^t)^{1/t} < (e^{t^2})^{1/t}$ Using properties of exponentiation again, this simplifies to: $(1+t) < e^t$ which is the desired inequality $e^{t}<(1+t)^{1+t}$ for $t>0$. Thus, we have proven that if $t>0$, then $e^{t}<(1+t)^{1+t}$.

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91影视

Combine to show that $$ \mathbf{1}+t0\) Show that if \(t>0\), then \(e^{t}<(1+t)^{1+t}\).

Short Answer

Step by step solution

Use the given inequality for non-negative x

Manipulate the inequality

Simplify and obtain the desired result

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