Minimum value of the expression

[gfd43tg]

Homework Statement

Problem is posted as image

Homework Equations

The Attempt at a Solution

Hello,

I am having some confusion over what is meant by 'type in the boxes the minimum value of the expression'. Does that mean take the derivative of the function? Or does that mean the value at which the function is a minimum? That would be setting them all to zero

a) $\underset \min{x} \hspace{0.05 in} x =$

$f(x) = x$
$f'(x) = 1$
$f'(x) = 0$ at the minimum
$1 \neq 0$
$\bar{x} = -\infty$

b) $\underset \min{x}\hspace{0.05 in}2x^2 =$

$f(x) = 2x^2$
$f'(x) = 4x$
$f'(x) = 0$ at the minimum
$4x = 0$
$\bar{x} = 0$c) $\underset \min{x} \hspace{0.05 in}x + 2x^2 =$

$f(x) = x + 2x^2$
$f'(x) = 4x + 1$
$f'(x) = 0$ at the minimum
$4x = -1$
$\bar{x} = -0.25$

d) $\underset \min{x}\hspace{0.05 in} 5 - x + 2x^2 =$

$f(x) = x + 2x^2$
$f'(x) = 4x - 1$
$f'(x) = 0$ at the minimum
$4x = 1$
$\bar{x} = 0.25$

 

[Mentallic]

It means to write the y value of the function that occurs at the minimum x value.
For the parabola

$$x+2x^2$$

the min occurs at $x=-0.25$ and hence the answer would be $$f(-0.25)=-0.25+2(-0.25)^2=-0.125$$

 

[gfd43tg]

Thanks, got it.