Problem

If $f$ has a local minimum value at $c$, show tha…

00:44

Problem 77 Hard Difficulty

Prove that the function
$$ f(x) = x^101 + x^51 + x + 1 $$
has neither a local maximum nor a local minimum.

Name: prove that the function fx x101 x51 x 1 has neither a local maximum nor a local minimum
Uploaded: 2019-11-06
Duration: 36 s
Channel: Numerade
Description: Prove that the function $$ f(x) = x^101 + x^51 + x + 1 $$ has neither a local maximum nor a local minimum.

Answer

$f(x)=x^{101}+x^{51}+x+1 \Rightarrow f^{\prime}(x)=101 x^{100}+51 x^{50}+1 \geq 1$ for all $x,$ so $f^{\prime}(x)=0$ has no solution. Thus, $f(x)$
has no critical number, so $f(x)$ can have no local maximum or minimum.

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Video Transcript

Let's first take the derivative. As we can see, we have positive coefficients and we have powers that are even 150 or both even. Therefore, we know it's always increasing something that's always increasing. Let's say it looks like this or looks like this. It doesn't have a minimum of Maxim because it's always constantly increasing.

Amrita B.

Topics

Derivatives

Differentiation

Volume

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 1

Maximum and Minimum Values

Problem

If $f$ has a local minimum value at $c$, show tha…

Problem 77 Hard Difficulty

Prove that the function
$$ f(x) = x^101 + x^51 + x + 1 $$
has neither a local maximum nor a local minimum.

Answer

$f(x)=x^{101}+x^{51}+x+1 \Rightarrow f^{\prime}(x)=101 x^{100}+51 x^{50}+1 \geq 1$ for all $x,$ so $f^{\prime}(x)=0$ has no solution. Thus, $f(x)$
has no critical number, so $f(x)$ can have no local maximum or minimum.

More Answers

Topics

Calculus: Early Transcendentals

Discussion

Top Calculus 2 / BC Educators

Grace H.

Catherine R.

Kayleah T.

Kristen K.

Calculus 2 / BC Bootcamp

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Watch More Solved Questions in Chapter 4

Video Transcript

Topics

Calculus: Early Transcendentals

Top Calculus 2 / BC Educators

Grace H.

Catherine R.

Kayleah T.

Kristen K.

Calculus 2 / BC Bootcamp

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Problem

If $f$ has a local minimum value at $c$, show tha…

Problem 77 Hard Difficulty

Prove that the function$$ f(x) = x^101 + x^51 + x + 1 $$has neither a local maximum nor a local minimum.

Answer

$f(x)=x^{101}+x^{51}+x+1 \Rightarrow f^{\prime}(x)=101 x^{100}+51 x^{50}+1 \geq 1$ for all $x,$ so $f^{\prime}(x)=0$ has no solution. Thus, $f(x)$has no critical number, so $f(x)$ can have no local maximum or minimum.

More Answers

Topics

Calculus: Early Transcendentals

Discussion

Top Calculus 2 / BC Educators

Grace H.

Catherine R.

Kayleah T.

Kristen K.

Calculus 2 / BC Bootcamp

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Watch More Solved Questions in Chapter 4

Video Transcript

Topics

Calculus: Early Transcendentals

Top Calculus 2 / BC Educators

Grace H.

Catherine R.

Kayleah T.

Kristen K.

Calculus 2 / BC Bootcamp

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Prove that the function
$$ f(x) = x^101 + x^51 + x + 1 $$
has neither a local maximum nor a local minimum.

$f(x)=x^{101}+x^{51}+x+1 \Rightarrow f^{\prime}(x)=101 x^{100}+51 x^{50}+1 \geq 1$ for all $x,$ so $f^{\prime}(x)=0$ has no solution. Thus, $f(x)$
has no critical number, so $f(x)$ can have no local maximum or minimum.