Math Is Fun Forum

jigbobber

Member

Registered: 2008-08-13

Posts: 4

help...having trouble

solve for x:

also this one

find the greatest and least value of

for -5≤x≤3

edit:found another confusing me..

if

show that

for all real x

Last edited by jigbobber (2008-08-13 01:32:51)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: help...having trouble

For the first one, change it to x³ + x² - 5x + 3 > 0 and try to find roots.

By inspection, x=1 is a root and so (x-1) will be a factor.
Using polynomial division turns it into (x-1)(x² + 2x - 3) > 0.
This in turn becomes (x-1)²(x+3) > 0.

So the graph would look like a cubic curve which starts negative, then gets to the x-axis at x=-3, goes positive for a bit, curves back and touches (but doesn't pass through) the x-axis again at x=1, and then stays positive after that.

The solution to the inequality would therefore be x > -3 and x ≠ 1.

For the second question, you differentiate the expression and look for roots. Then check any roots you find against the original expression, as well as the two end-points. The greatest and least values will be one of these.

Differentiating x³ - 12x - 5 gives 3x² - 12 = 3(x² - 4) = 3(x+2)(x-2).
It is therefore equal to 0 at ± 2.

So you check the function's value at -5, -2, 2 and 3 and choose the least and greatest values of those.

f(-5) = -70
f(-2) = 11
f(2) = -21
f(3) = -14

From that, it's easy to see that the function is lowest when x=-5 and highest when x=-2.

The last question should be easy enough if you know the quotient rule.
Differentiate the expression using that, and you should end up with a fraction whose numerator and denominator are clearly always positive. Therefore, the derivative will also always be positive.

---

Why did the vector cross the road?
It wanted to be normal.

jigbobber

Member

Registered: 2008-08-13

Posts: 4

Re: help...having trouble

thanks for the fast reply mathsyperson. you really helped.