what are the stationary points of f(x)=20+16x-x^2/4+x^2?

Question

I have used the quotient rule to differentiate f(x)=20+16x-x^2/4+x^2. This gives the answer f'(x)=16(4-3x-x^2)/(4+x^2)^2. Now I have to find the stationary points of f(x). I understand in principle how to do this but am stuck because f(x) is a fraction, and I am not sure how to manipulate it into a one line sum

Answer 1

f'(x)=16(4-3x-x^2)/((4+x^2)^2
i presume this is correct
if the derivative should vanish the numerator should vanish
that is x^+3x-4=0
(x+4)(x-1)=0
so x=-4 or 1

Answer 2

The stationary points occur when f'(x)=0 (which from your question I think you understand).
So now you need to find what x is when 16(4-3x-x^2)/(4+x^2)^2=0
If a/b=0 the only solution is a=0. Therefore we can ignore everything except 4-3x-x^2=0. If you're doing calculus, you should be able to find the answers to that fairly easily, so I'll leave it to you.

Answer 3

By rule, a fraction will only equal 0 (which is what you're trying to get to) if the numerator = 0. Therefore, you want to make 16(4-3x-x^2)=0. Start from there, and use whatever other methods you know how to!

Answer 4

f'(x)=0+16-x/2+2x=16+3x/2=slope=0

a stationary point is a point on the graph of a function where the tangent to the graph is parallel to the x-axis

3x/2=16

x=16 . 2/3=32/3

Answer 5

(1,7) and (-4,-3)

For stationary points, the gradient = 0, so f'(x)=0.

f'(x)=16(4-3x-x^2)/(4+x^2)^2
0 = 16(4-3x-x^2)
x^2 + 3x - 4 = 0
(x-1)(x+4) = 0
x = 1 or -4

f(1) = 35/5 = 7
f(-4) = -60/20 = -3

(Note: We ignore the denominator of 4+x^2 because it will always be a positive number and can never be zero.)

So the stationary points are (1,7) and (-4,-3).

Answer 6

x^2 u recommend? x^2 + 16x +sixty 9 = 6 get rid of 6 from sixty 9 and six from after the equivalent signal. x^2 + 16x + sixty 3 = 0 use quadratic formula [it really is: -b +/- (sqrt b^2 - 4ac) / 2a] - 16 +/- (sqrt 16^2 - 4 * a million * sixty 3) / 2 -16 +/- (sqrt 256 - 252) /2 -16 +/- (sqrt 4)/2 [-16 +/- (2)] / 2 -7 or -9 is the answer

Answer 7

f(x) = (20 + 16x - x^2) / (4 + x^2)

f'(x) = (16-2x)/(4+x^2) - (20+16x-x^2)2x/(4+x^2)^2

if f'(x)=0 then

(16-2x)(4+x^2) - (20+16x-x^2)2x = 0

or 64 - 8x + 16x^2 - 2x^3 - 40x - 32x^2 + 2x^3 = 0

or -16x^2 - 48x + 64 = 0 or x^2 + 3x - 4= 0 or x=1 or x= - 4

note: learn the rule for differentiating fractional functions

f ' (u/v) = (v f ' (u) - u f ' (v) ) / v^2

Answer 8

f(x)=20+16x-x^2/4+x^2
f'(x) = [(2x)(20+16x-x^2) - (16-2x)(4+x2)] / (4 +x^2)^2
f'(x) = [(20x+32x^2-2x^3) - (64+16x^2-8x-2x^3)] / (4 +x^2)^2
=[(20x+32x^2-2x^3) - (64+16x^2-8x-2x^3)] / (4 +x^2)^2
On simplifying we get
f'(x) =16(x^2 +3x -4)/(x2+4)^2
to find the critical points put numerator=0 and Denominator =0
x^2 +3x -4=0 0r (x-4)(x+1)=0 or x =4 and x=-1
Sine x^2+4=0 0r x=+/-(2i) which have imaginary points
Hence the critical points are -1,4

Answer 9

That is a hard sum. For me anyway. I'm aware that x equals 20, but what do'es the little triangle pointy thing mean?

Answer 10

I was going to say that but he beat me to it.