Grade 12 Calculus Problem - Differentiation and Division

In summary, the conversation discusses a question about a polynomial function with certain conditions for a real value a. The options are given and the conversation includes hints and explanations to narrow down the possible answers. Eventually, the correct answer is determined to be option E) p(x) is divisible by x^2-2ax+a^2.
  • #1
G-S
10
0

Homework Statement



Let p be a polynomial function with p(a)= 0 =p'(a) for some real a. Which of the following must be true?

A) p(x) is divisible by x+a
B) p(x) is divisible by x^2+a^2
C) p(x) is divisible by x^2-a^2
D) p(x) is divisible by x^2+2ax+a^2
E) p(x) is divisible by x^2-2ax+a^2

Homework Equations


The Attempt at a Solution



I'm honestly beyond stumped for this one. Having a hard time finding out where to start on this problem.
 
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  • #2
What does it mean for a polynomial (or any function) if p(x) = 0 at x=a ?

What does it mean for a polynomial, if p'(x) = 0 at x=a ?

Put the answers together.
 
  • #3
Hint: p(x) is a polynomial and p(a)=0 means a is a root of the polynomial. If you were to factor p(x), what factor do you know has to be there if a is a root?
 
  • #4
Thanks for the replies.
So this is the conclusion I've come to so far,

p(x)=0 at x=a
p'(x)=0 at x=a
a is a root of the polynomial

So,
x2-a2=0
x2=a2
x=a

Therefore, C) must be true.

Am I in the right direction?
 
  • #5
No, unless you know that -a is also a root.

What is true about a function if it's first derivative is zero at some value of x ?
 
  • #6
Sorry SammyS but the only thing that comes to mind is the function is a constant. Either that or the first step to find critical points.

After looking at the problem again I've come to the conclusion that in order to fulfill the condition p(a)=0 only C) and E) can be true.

Considering that we only know that a is a root (and not -a) E) seems like the only one that fulfills the conditions.

0=x2-2ax+a2
0=(x-a)2
0=x-a
x=a

Taking stabs in the dark here.
 
  • #7
G-S said:
After looking at the problem again I've come to the conclusion that in order to fulfill the condition p(a)=0 only C) and E) can be true.

Considering that we only know that a is a root (and not -a) E) seems like the only one that fulfills the conditions.

You have narrowed down the list of choices to just E, so E must be the answer. Is there anything more you need to know?
 
  • #8
e(ho0n3 said:
You have narrowed down the list of choices to just E, so E must be the answer. Is there anything more you need to know?

I'm trying to get confirmation as to whether the answer and my thought process is correct and if not, how one would approach this question.
 
  • #9
G-S said:
Sorry SammyS but the only thing that comes to mind is the function is a constant. Either that or the first step to find critical points.

...
And, in general, why is it that you look for critical points?

Also, what is the slope of a constant function? (I'm assuming that this is not a constant polynomial.)
 
  • #10
SammyS said:
And, in general, why is it that you look for critical points?

Also, what is the slope of a constant function? (I'm assuming that this is not a constant polynomial.)

To find max/min values. And the slope of a constant function is 0.
 
  • #11
More to the point, what is the slope of the line tangent to the polynomial at x = a, if p'(a) = 0 ?

This is the question I should have asked in my previous post.
 
  • #12
The slope of the tangent line is 0 if p'(a)=0.
 
  • #13
The behavior of a polynomial in the neighborhood of one of its zeros is due mainly to the factor which 'causes' the zero, and the multiplicity of that zero.

So, in the neighborhood of x=a, the polynomial, p(x), behaves like ±(x - a)n, where n ≥ 2 .

How do we know it's not like ±(x - a)1 ?
 
  • #14
SammyS said:
How do we know it's not like ±(x - a)1 ?

Because it's not one of the options? Haha, I haven't got a clue.
 
  • #15
The derivative of (x - a) ≠ 0 at x = 1 .
 
  • #16
SammyS said:
The derivative of (x - a) ≠ 0 at x = 1 .
I presume you mean at x= a.
 
  • #17
Yes. Thanks for the correction.
 

Related to Grade 12 Calculus Problem - Differentiation and Division

1. What is differentiation and why is it important in calculus?

Differentiation is a mathematical operation used to find the rate of change of a function. In calculus, it is important because it allows us to calculate the slope of a curve at any given point, which is crucial in understanding the behavior of functions and solving many real-world problems.

2. What is the difference between differentiation and division?

Differentiation is a mathematical operation that involves finding the derivative of a function, while division is a basic arithmetic operation used to find the quotient of two numbers. In calculus, differentiation is used to find the rate of change of a function, while division is used to find the ratio between two quantities.

3. How do you differentiate a polynomial function?

To differentiate a polynomial function, we use the power rule, which states that the derivative of a term with an exponent n is equal to n times the coefficient of the term, multiplied by the variable raised to the power of n-1. We then add up the derivatives of each term to find the overall derivative of the polynomial function.

4. Can you use differentiation to find the maximum or minimum of a function?

Yes, we can use differentiation to find the maximum or minimum of a function. We do this by finding the critical points of the function, which are the points where the derivative of the function is equal to zero. We then use the second derivative test to determine whether these points are a maximum or minimum.

5. What is the chain rule and how is it used in differentiation?

The chain rule is a rule used to find the derivative of a composite function. It states that to find the derivative of a composite function, we multiply the derivative of the outer function by the derivative of the inner function. This allows us to differentiate more complex functions by breaking them down into simpler parts.

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