interlanken-fall
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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.
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Callicious
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They're asking you to constrain p given that the gradient of f(x) is positive within [-1,1].

Differentiate f(x), i.e. find \frac{df(x)}{dx}, and try to think about what is necessary for this quantity to be positive subject to x\in[-1,1].
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interlanken-fall
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(Original post by Callicious)
They're asking you to constrain p given that the gradient of f(x) is positive within [-1,1].

Differentiate f(x), i.e. find \frac{df(x)}{dx}, and try to think about what is necessary for this quantity to be positive subject to x\in[-1,1].
this is the answer
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
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Callicious
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(Original post by interlanken-fall)
this is the answer
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 x=-1 you have the condition that p\geq{2}. You also know that for the upper limit x=1 you get the condition of p\geq{-2}.

You need both these conditions to satisfy your initial problem. If p\geq{2}, you also satisfy the condition of p\geq{-2} 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;Name:  1241251251.PNG
Views: 6
Size:  51.3 KB

Sorry about the awful artwork, but you get the drift.
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interlanken-fall
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(Original post by Callicious)
You've ascertained that for the lower limit of your domain x=-1 you have the condition that p\geq{2}. You also know that for the upper limit x=1 you get the condition of p\geq{-2}.

You need both these conditions to satisfy your initial problem. If p\geq{2}, you also satisfy the condition of p\geq{-2} 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;Name:  1241251251.PNG
Views: 6
Size:  51.3 KB

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
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Callicious
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(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
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