-
Best of luck to the class of 2024 for their HSC exams. You got this! Let us know your thoughts on the HSC exams here -
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page
calculus qn (1 Viewer)
- Thread starter bos1234
- Start date
P
pLuvia
Guest
a) Find f'(x), equate it to zero. Then solve for x and you should realise that the discriminant is negative hence there are no roots to the derivative hence no stationary points for f(x)
b) After you find f'(x), find the limits as x approaches - and + infinity and you should observe f'(x) will be positive for both of the limits hence f(x) is strictly increasing for all x
c) Well if f(x) is strictly increasing what you can you observe about the graph and its intercepts?
b) After you find f'(x), find the limits as x approaches - and + infinity and you should observe f'(x) will be positive for both of the limits hence f(x) is strictly increasing for all x
c) Well if f(x) is strictly increasing what you can you observe about the graph and its intercepts?
jb_nc
Google "9-11" and "truth"
- Joined
- Dec 20, 2004
- Messages
- 5,391
- Gender
- Male
- HSC
- N/A
f(x) = 2x³ - 3x² + 5x + 1
f'(x) = 6x² - 6x + 5
For st. pts. f'(x) = 0 into quad. formula which gives a complex solution i.e. 4ac > b² so in terms of 2 unit mathematics there is no st. pts.
b) Examine the limit of the function as x→-∞ which is +∞ , x→0 which = 5 and x→∞ which is +∞. If we examine the function itself we can see the 6x² part will increase faster than 6x and the squared will convert any negative number into a positive one. The range of f'(x) is [5, ∞) so that means f'(x) > 0 for all real values of x.
c) Eh, I'm not sure what "deduce" means for f(x) but there is a formula for the discriminate of a cubic equation which is: ∆ = 4b³d - b²c² + 4ac³ - 18abcd + 27a²d² (lol) where the polynomial is in the form ax³ + bx² + cx + d = 0. If you put the coefficients of f(x) into the formula you get a number >0 which means that it has one real root (and a pair of complex conjugates roots).
I posted mine before pluvia but deleted it because I don't like that edit thing at the bottom, it's also better than his (read mine).
f'(x) = 6x² - 6x + 5
For st. pts. f'(x) = 0 into quad. formula which gives a complex solution i.e. 4ac > b² so in terms of 2 unit mathematics there is no st. pts.
b) Examine the limit of the function as x→-∞ which is +∞ , x→0 which = 5 and x→∞ which is +∞. If we examine the function itself we can see the 6x² part will increase faster than 6x and the squared will convert any negative number into a positive one. The range of f'(x) is [5, ∞) so that means f'(x) > 0 for all real values of x.
c) Eh, I'm not sure what "deduce" means for f(x) but there is a formula for the discriminate of a cubic equation which is: ∆ = 4b³d - b²c² + 4ac³ - 18abcd + 27a²d² (lol) where the polynomial is in the form ax³ + bx² + cx + d = 0. If you put the coefficients of f(x) into the formula you get a number >0 which means that it has one real root (and a pair of complex conjugates roots).
I posted mine before pluvia but deleted it because I don't like that edit thing at the bottom, it's also better than his (read mine).
P
pLuvia
Guest
Just prove it using the first derivative, if the gradient is increasing for a certain domain then within that domain the function is strictly one-to-one hence increasing which must mean that it will not have a turning point in that interval