# differentiation question A1 math

Watch
#1
Given that the function f(x) = x2 + px is increasing on the interval [−1, 1], find
one possible value for p. (2 marks)
b State with justification whether this is the only possible value for p.
0
3 days ago
#2
They're asking you to constrain given that the gradient of is positive within [-1,1].

Differentiate , i.e. find , and try to think about what is necessary for this quantity to be positive subject to .
1
#3
(Original post by Callicious)
They're asking you to constrain given that the gradient of is positive within [-1,1].

Differentiate , i.e. find , and try to think about what is necessary for this quantity to be positive subject to .
f(x) = x2 + px
f′(x) = 2x + p ≥ 0 when −1 ≤ x ≤ 1
When x = −1, f′(x) = −2 + p ≥ 0, so p ≥ 2
So for f′(x) ≥ 0, p ≥ 2 e.g. p = 3
When x = 1, f′(x) = 2 + p ≥ 0, so p ≥ −2
However, p ≥ 2 to work with x = −1.
b Using the proof from part a, any value
p ≥ 2 will work.

the bit i don't get is why is p ≥ 2 then p ≥ −2
0
3 days ago
#4
(Original post by interlanken-fall)
f(x) = x2 + px
f′(x) = 2x + p ≥ 0 when −1 ≤ x ≤ 1
When x = −1, f′(x) = −2 + p ≥ 0, so p ≥ 2
So for f′(x) ≥ 0, p ≥ 2 e.g. p = 3
When x = 1, f′(x) = 2 + p ≥ 0, so p ≥ −2
However, p ≥ 2 to work with x = −1.
b Using the proof from part a, any value
p ≥ 2 will work.

the bit i don't get is why is p ≥ 2 then p ≥ −2
You've ascertained that for the lower limit of your domain you have the condition that . You also know that for the upper limit you get the condition of .

You need both these conditions to satisfy your initial problem. If , you also satisfy the condition of and hence this must be your answer.

Whenever I hit these kinds of problems, I drew a number line and dashed my limits, like so;

Sorry about the awful artwork, but you get the drift.
1
#5
(Original post by Callicious)
You've ascertained that for the lower limit of your domain you have the condition that . You also know that for the upper limit you get the condition of .

You need both these conditions to satisfy your initial problem. If , you also satisfy the condition of and hence this must be your answer.

Whenever I hit these kinds of problems, I drew a number line and dashed my limits, like so;

Sorry about the awful artwork, but you get the drift.
thank you! so much so for the 2 functions
-2+p≥ 0 p≥2
2+p≤0 p≥-2
all the values of p≥2 satisfy both functions so thus we choose it? is this right?
Last edited by interlanken-fall; 3 days ago
1
3 days ago
#6
(Original post by interlanken-fall)
thank you! so much so for the 2 functions
-2+p≥ 0 p≥2
2+p≤0 p≥-2
all the values of p≥2 satisfy both functions so thus we choose it? is this right?
They're inequalities, but you're right. Glad it worked out
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### What would make a great online university open day?

Virtual campus tour (68)
19.05%
Virtual accommodation tour (51)
14.29%
Find out about sports clubs, societies and facilities (17)
4.76%
Video content about the local area (3)
0.84%
Webinars with lecturers (11)
3.08%
Taster lectures or seminars (76)
21.29%
Speak to current students studying my course (81)
22.69%
Speak to current students about the uni in general (16)
4.48%
Fun online activities or experiences (13)
3.64%
Connecting with careers services or employers (8)
2.24%
Info about student wellbeing and support services (6)
1.68%
Something else (let us know in the thread!) (7)
1.96%