Prove y^5+y^2-7y+5≥0 for y≥1

[evagelos]

Homework Statement

Prove that: $$y^5+y^2-7y+5\geq 0$$ ,for all $$y\geq 1$$

Homework Equations

The Attempt at a Solution

$$y^5\geq 1$$ and $$y^2\geq 1$$ => $$y^5+y^2\geq 2$$.

Also $$-7y+5\leq -2$$ , and then?

 

[Mark44]

Homework Statement

Prove that: $$y^5+y^2-7y+5\geq 0$$ ,for all $$y\geq 1$$

Homework Equations

The Attempt at a Solution

$$y^5\geq 1$$ and $$y^2\geq 1$$ => $$y^5+y^2\geq 2$$.

Also $$-7y+5\leq -2$$ , and then?

If f(y) = y⁵ + y² - 7y + 5, note that f(1) = 0.

Look at f'(y) to determine where f is increasing and decreasing for y >= 1.

 

[evagelos]

If f(y) = y⁵ + y² - 7y + 5, note that f(1) = 0.

Look at f'(y) to determine where f is increasing and decreasing for y >= 1.

O.K

f'(y)=$$5y^4+2y-7\geq 0$$ for $$y\geq 1$$ .

But how can this effect our f(y) ??

 

[Mark44]

f'(y) gives the slope at a point (y, f(y)) on the graph of f. If f'(y) > 0, the graph of f is increasing. If f'(y) < 0, the graph of f is decreasing.

We know that f(1) = 0. Is the graph of f going up or down from there?

BTW, this seems to be a calculus problem, so it should have been posted in the Calculus & Beyond section, not the Precalculus section.

 

[ehild]

Write f(y) in terms of x=y-1≥0.

ehild

 

[evagelos]

f'(y) gives the slope at a point (y, f(y)) on the graph of f. If f'(y) > 0, the graph of f is increasing. If f'(y) < 0, the graph of f is decreasing.

We know that f(1) = 0. Is the graph of f going up or down from there?

BTW, this seems to be a calculus problem, so it should have been posted in the Calculus & Beyond section, not the Precalculus section.

you can have f(y)<0 and f'(y)>=0 ,so we cannot get a contradiction

 

[Tedjn]

If the graph is always increasing, and its value at the leftmost point (y=1) is 0, then can it be negative to the right (y>1)? You can also follow ehild's suggestion to get a more direct answer.

 

[evagelos]

Write f(y) in terms of x=y-1≥0.

ehild

Like this?

$$(x+1)^5+(x+1)^2-7(x+1)+5$$

 

[Tedjn]

Expand.

 

[Mark44]

Like this?

$$(x+1)^5+(x+1)^2-7(x+1)+5$$

This is really the long way around. You have the derivative -- f'(y) = 5y⁴ + 2y - y. It's a very simple matter to show that f'(y) >= 0 for y >= 1, hence the graph is increasing for y >= 1, and you're pretty much done.

 

[Dickfore]

Factor the polynomial. Mathemtica gives:

$$y^{5} + y^{2} -7y+5 = (y - 1)^{2} \, g_{3}(y)$$

where $g_{3}(y)$ is a polynomial of 3rd digree. What are the extremal values of the polynomial $g_{3}(y)$ in the interval $y \ge 1$?

 

[ehild]

This is really the long way around.

It is not that difficult to expand, knowing the coefficients of (x+1)⁵ from Pascal's Triangle.

(x+1)⁵=x⁵+5x⁴+10x³+10x²+5x+1

ehild