Geometric applications of calculus help! (1 Viewer)

[meg024]

1. Find all values of x for which the curve f(x)=x^3-3x+4 is decreasing
2. Find the domain over which the curve y=x^3+12x^2+45x-30 is increasing

I got the right numbers for both but my greater than, less than signs are all mixed up so if anyone could explain the signs part that would be great! thankyouu

 

[TimeGoesTooFast]

1. f(x)= x^3-3x+4
f'(x)=3x^2-3
For decreasing: 3x^2-3< 0
therefore: x^2-1<0 (draw graph)
-1< x <1


2. f(x)=x^3+12x^2+45x-30
f'(x)=3x^2+24x+45
For increasing: 3x^2+24x+45>0
(3x+9)(x+5)>0 (draw graph)
x <-3 and x >-5

If you graph the inequality on a number line, you'll be able see it. Hope this helps.

 

[meg024]

thankyouu!

 

[meg024]

Just another question
Show the curve y=x^3-3x^2+27x-3 is monotonic increasing for all values of x?

 

[TimeGoesTooFast]

Don't get throw off by the phrase "monotonically increasing". It's just asking "for what values of x is the curve always increasing".
y'=3x^2-6x-27
3x^2-6x-27>0 (draw graph)
(3x+3)(x-9)>0
x<-1 and x>9.

 

[meg024]

In the answers of my text book it does a>0 and then b^2-4ac to get the answer and the answer doesn't have -1 or 9 it just says therefore it is monotonic increasing. Because wouldn't the greater and less than signs mean it was only increasing outside or -1 and 9 rather than monotonic increasing?

 

[HeroicPandas]

In the answers of my text book it does a>0 and then b^2-4ac to get the answer and the answer doesn't have -1 or 9 it just says therefore it is monotonic increasing. Because wouldn't the greater and less than signs mean it was only increasing outside or -1 and 9 rather than monotonic increasing?

 

[meg024]

thanks

 

[meg024]

Can someone please help me with differentiating these:
a) P=2x+ 50 over x
b) A=h^2-2h+5 all over 8
c)V=40r-piR^3

thankyouu

 

[nightweaver066]

Differentiating all these just requires you to use this:

 

[girlworld_club]

1. f(x)= x^3-3x+4
f'(x)=3x^2-3
For decreasing: 3x^2-3< 0
therefore: x^2-1<0 (draw graph)
-1< x <1


2. f(x)=x^3+12x^2+45x-30
f'(x)=3x^2+24x+45
For increasing: 3x^2+24x+45>0
(3x+9)(x+5)>0 (draw graph)
x <-3 and x >-5

If you graph the inequality on a number line, you'll be able see it. Hope this helps.

how would you draw the graph in a test?

 

[girlworld_club]

1. f(x)= x^3-3x+4
f'(x)=3x^2-3
For decreasing: 3x^2-3< 0
therefore: x^2-1<0 (draw graph)
-1< x <1


2. f(x)=x^3+12x^2+45x-30
f'(x)=3x^2+24x+45
For increasing: 3x^2+24x+45>0
(3x+9)(x+5)>0 (draw graph)
x <-3 and x >-5

If you graph the inequality on a number line, you'll be able see it. Hope this helps.

p.s your answer is wrong. X>-3, and x<-5. And when i graphed it it did not help.

 

[HeroicPandas]

p.s your answer is wrong. X>-3, and x<-5. And when i graphed it it did not help.

Why doesn't the graph help?

When you graph (on just the x-axis) and x=/=5 x=/= -3

Look at the graph when x <-5, its positive
Look at the graph when -5 < x < -3, its negative
Look at the graph when x > -3, its positive

We want positive, therefore we take the positive inequalities (as you have found)

 

[girlworld_club]

yes i understand, but the graph it not necessary and over complicates it. What i simply do is get the two points in this case x=-5, and -3. And pick a point in between. I sub this point into the derived equation and if it satisfies then i know the answer. What i don't get is why do we derive the original question? to find points in which it increases?

 

[HeroicPandas]

yes i understand, but the graph it not necessary and over complicates it. What i simply do is get the two points in this case x=-5, and -3. And pick a point in between. I sub this point into the derived equation and if it satisfies then i know the answer. What i don't get is why do we derive the original question? to find points in which it increases?

That method is great too!
The derivative shows when the curve is increasing (f'(x) > 0) & decreasing (f'(x) < 0) (values of x)
We introduce an inequality to find values where the curve is inceasing or decreasing

f'(x) = 0 finds stationary pts (a point on the curve where the gradient is zero)