help appreciated with these questions! (1 Viewer)

[ronaldinho]

its from the year 11 cambridge book..page 360-361

(a) Prove that f(x) = 2x^3 - 3x^2 +5x + 1 has no stationery pts

(b) show that f'(x) > 0 for all values of x, and hence that the function is always increasing

(c) Deduce that the eqn of f(x) = 0 has only one real root

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13
(a) Prove that y= -x^3 + 2x^2 -5x + 7 is deacreasing for all values of x

(b) Hence deduce the number of solns to the eqn 7 - 5x + 2x^2 - x^3 = 0

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A function f(x) has derivative f'(x) = -x(x+2)(x-1)

a)Draw a graph of y = f'(x), and hence establish where f(x) is increasing, decreasing and stationery.

could u plz find where i made the mistake because when i draw the graph of y = f'(x) its different from the book...

 

[BIRUNI]

(a) dy/dx=6x^2-6x+ 5 delta<0

(b) 6>0 and discriminant is negative so your derivative graph is always above the x-axis
(c) since there is no maxima and minima point and dy/dx>0 for all real x that means the curve is monotonic increasing and cuts the x-axis once does not come down to cut the x-axis again

 

[BIRUNI]

13. dy/dx=-3x^2+ 4x-5 delta is less than o and -3 is less than o so the derivative graph is always below the x axis that meants it is monotonic decreasing
(b) as x goes to negative infinity, y goes to positive infinity so when you are looking at the curve from left side, the curve comes down from above the ax-axis, cut the x-axis at one point and it decreases all the time and never comes up to cut the x-axis again

 

[ronaldinho]

is there any toher way of doing this..?

i done it like this but it might not b enuf.. i dont know how they makr these questions

(a)f'(x) = 6x^2 -6x + 5
at stationery pts dy/dx = 0

6x^2 - 6x + 5 = 0
then i used the formula -b=-square root b^2 - 4ac/2a.. etc.. and no soln..l therefore no stationery pts [right or wrong?]

(b) i dont really know how to show this other than sub in values....

(c) dont know

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13 - have some idea.... but i dot know how to set out these workings..



anyways thanks for ur help man...but ur doing it a bit advanced way i think...

 

[airie]

How did you reason that there are no solutions to the equation in section (a)? If you put the values into the quadratic formula, you would've gotten a negative value under the square root sign. You could've just used the fact that this discriminant is negative to reason that there are no real roots to the equation (6x^2-6x+5=0), thus 6x^2-6x+5 cannot equal to zero (as the x-coordinates of all points on the f'(x) graph are real) ie. f'(x) never equal zero, there are no stationary points on the graph of f(x). That's BIRUNI's approach.

As for (b), there are an infinitive number of values that x can take, so you can't just sub in a few values and say that since it doesn't work for those it won't work for all values of x. BIRUNI drew out the graph of f'(x) - a parabloa, and since it does not touch the x-axis (as 6x^2-6x+5 does not equal to zero for all values of x), and the coefficient of x^2 is positive, the graph is a concave up parabola that is entirely above the x-axis ie. the value of 6x^2-6x+5 is positive for all x. You could also complete the square: 6x^2 - 6x + 5 = 6(x - 1/2)^2 + 7/3, and since 6(x - 1/2)^2 is greater that or equal to zero for all x, 6x^2 - 6x + 5 = 6(x - 1/2)^2 + 7/3 > 0 for all x, as required.

You now know that the derivative of f(x) is positive for all values of x, which means that it is increasing over the whole domain ie. there are no two different values of x for which f(x) takes the same value as f(x) is ever increasing. So if it does cut the x-axis (in this case it does, since there are no horizontal asymptotes at all), it will not cut the axis in any other place.

A function f(x) has derivative f'(x) = -x(x+2)(x-1)

a)Draw a graph of y = f'(x), and hence establish where f(x) is increasing, decreasing and stationery.

could u plz find where i made the mistake because when i draw the graph of y = f'(x) its different from the book...

How did your graph look like, and how did you go about drawing it?

I've drawn one in Paint, the graph of f'(x) should look something like it, where f'(x) has roots at x = 0, -2, 1, and the coefficient of the x^3 is negative after expansion - you can tell by the negative sign at the very front. As you know, f(x) is increasing when f'(x) is positive, f(x) is decreasing when f'(x) is negative, and f(x) is stationary when f'(x) is zero. Looking at the graph of f'(x), it is positive when x < -2 and when 0 < x < 1, and negative when -2 < x < 0 and x > 1.

Therefore, f(x) is increasing when x < -2 and when 0 < x < 1,
decreasing when when -2 < x < 0 and x > 1,
and stationary when x = 0, -2, or 1.

Hope that helps

 

[ronaldinho]

ok thanks for the helps.. i got it now...
when they say explain and deduce can i put it in words like uve done.. or do i need to show it thru working?

do u think this is 3 unit or 2 unit standard question?

 

[airie]

Uhh...I guess you can do either, really. As long as you show your marker how did you get your answers, so that they know you didn't just have a lucky guess or something

I probably wouldn't know if it's of 2u or 3u standards, having not seen a 2u textbook myself Would you be able to do a similar question, now that you get it? If so, you should do fine

 

[ronaldinho]

Uhh...I guess you can do either, really. As long as you show your marker how did you get your answers, so that they know you didn't just have a lucky guess or something

I probably wouldn't know if it's of 2u or 3u standards, having not seen a 2u textbook myself Would you be able to do a similar question, now that you get it? If so, you should do fine

yeah i done the next question which is similar to the ones i asked..i can do them now!

 

[SoulSearcher]

ok thanks for the helps.. i got it now...
when they say explain and deduce can i put it in words like uve done.. or do i need to show it thru working?

do u think this is 3 unit or 2 unit standard question?

Harder 2 unit, probably.