Urgent Help:
if f(x)=x^4+ax^2+b has a stationary point (1,4) find:

a and b

insert (1,4) in the equation

4=1^4+a(1)^2+b
4=1+a+b
a+b=3
a=3-b

how to find the value of a and b?

Question

Urgent Help:
if f(x)=x^4+ax^2+b has a stationary point (1,4) find:

a and b

insert (1,4) in the equation

4=1^4+a(1)^2+b
4=1+a+b
a+b=3
a=3-b

how to find the value of a and b?

aum · Answer

You then take the derivative, equate it to zero, put x = 1 and find 'a'.

anonymous · Answer

the derivative of f(x)?

aum · Answer

This is a calculus question, right?

anonymous · Answer

yes

anonymous · Answer

the stationary point is f'(x)=0

aum · Answer

stationary point means the derivative is zero at that point.

anonymous · Answer

i mean the gradient

aum · Answer

Yes, the slope is zero at that point.
If f'(c) = 0, then c is a stationary point.

anonymous · Answer

y'= 4x^3+0*x^2+2x*a+0

anonymous · Answer

y'= 4x^3+2xa

anonymous · Answer

4(1)^3+2(1)*a = 0

aum · Answer

correct.  solve for 'a'

anonymous · Answer

am i doing it right?

anonymous · Answer

great

anonymous · Answer

4 +2a = 0
2a=-4
a=-2

aum · Answer

yes.  put it in the other equation and find b.

anonymous · Answer

a=3-b
-2=3-b
b=5

anonymous · Answer

thank you so much for your help :)

anonymous · Answer

i really appreciate it

aum · Answer

You are welcome. I plotted f(x) here after putting in the values for 'a' and 'b'. As you can see, point (1,4) is on the curve and the tangent at (1,4) will be parallel to the x-axis with a slope of zero: https://www.wolframalpha.com/input/?i=x^4-2x^2%2B5&lk=4&num=1

anonymous · Answer

one more question... why let x=1 to find a?

anonymous · Answer

because i used x=1 in both equation f(x) and f'(x)

aum · Answer

(1,4) is a point on the curve.
So the point should satisfy f(x).
That is, 4 = f(1)

(1,4) is a stationary point  which implies the slope of f(x) is zero at (1,4)
So f'(1) = 0

anonymous · Answer

Thanks :)

aum · Answer

If f'(c) > 0, the slope is positive at x = c and the function is increasing at x = c
If f'(c) < 0, the slope is negative at x = c and the function is decreasing at x = c
If f'(c) = 0, the slope is zero.  The function is neither increasing nor decreasing at x = c.  That is why it is called a "stationary point".

anonymous · Answer

if x=2 is the stationary point then what will y be?

anonymous · Answer

@aum

aum · Answer

Did you find x = 2 is a stationary point or are you just asking for some random x value?

aum · Answer

Either way, if you know x, all you have to do to find y is to put it in the function and evaluate f(x) when x = 2.

anonymous · Answer

there is another question that states if x^3+ax^2+bx has stationary points at x=2 and x=-4/3

find a and b?

anonymous · Answer

f'(2)=0?

anonymous · Answer

f'(-4/3)=0

aum · Answer

Exactly.  You will get 2 equations and you can solve for the 2 unknowns: a and b.

anonymous · Answer

sweet ill try solve it and correct me if im wrong please

aum · Answer

okay.

anonymous · Answer

i got a different answer to what is in the textbook

aum · Answer

a = -1  and  b = -8

anonymous · Answer

your answers are correct

anonymous · Answer

how were you able to answer it?

aum · Answer

f(x) = x^3+ax^2+bx
f'(x) = 3x^2 + 2ax + b

f'(2) = 3(2)^2 + 2a(2) + b = 0
12 + 4a + b = 0
4a + b = -12  ----- (1)

f'(-4/3) = 3(-4/3)^2 + 2a(-4/3) + b = 0
3(16/9) - 8a/3 + b = 0
16/3 - 8a/3 + b = 0
multiply by 3 to get rid of fractions:
16 - 8a + 3b = 0
-8a + 3b = -16  ----- (2)
Multiply (1) by 2 and add it to (2):
  8a + 2b = -24
-8a + 3b = -16
add
           5b = -40;  b = -8.  Put this in (1):
4a - 8 = -12
4a = -4
a = -1

a = -1, b = -8

anonymous · Answer

gosh i did a stupid mistake i inserted the value of x in f(x) instead of f'(x)

anonymous · Answer

now it makes sense

aum · Answer

you got to be careful

anonymous · Answer

exactly i shouldnt rush

anonymous · Answer

i understand what is a local maximum and a local minimum but what do they mean by "a negative point of inflection"?

aum · Answer

y = x^3+ax^2+bx a = -1, b = -8 y = x^3 - x^2 - 8x If you plot it, you will notice the stationary points are indeed at x = 2 and x = -4/3 http://prntscr.com/45dtae

aum · Answer

"point of inflection" is when f'(x) = 0 but the point is neither a maximum not a minimum.  Such points will have the second derivative to be zero also.
"c"  could be a point of inflection if f'(c) = 0 and f''(c) = 0.

If c is negative, I suppose you can call it a negative point of inflection.

anonymous · Answer

thank you again .. your explanations have been really helpful

aum · Answer

You are welcome.

I have not heard the term "negative point of inflection" and so you may want to consult your book.  But the above description is valid for "point of inflection"