Pre-cal: Find a value of k such that the function has one zero?

Question

f(x)= 1/5 (x^5 - 20x^3 + 64x) + k

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Find a value of k such that the function only has one zero

Answer 1

First, have the function be equal to zero:

0 = 1/5 (x^5 - 20x^3 + 64x) + k

Take out the extra x:

0 = 1/5x (x^4 - 20x^2 + 64) + k

Now, what you want to do is complete the square because then when you factor what is in the parentheses, you want the two new quantities to be the same [ex. (x+1)(x+1)].  If you looked at the equation without the 1/5x, you'll see to make it a square of (x-10), then 64 needs to be 100.  So now you need to try and add another 36 to this equation.  By putting back in 1/5 (and even plug back in the x) you'll see that k cannot be exactly 36.

0 = 1/5x^5 - 4x^3 + 64/5x + k

Now, k must be 36/5x.  If you check this answer by replacing k with this new answer, it should all work out.

0 = 1/5x^5 - 4x^3 + 64/5x + 36/5x

0 = 1/5x (x^4 - 20x^3 + 100x)

0 = 1/5x (x-10)^2

Answer 2

dude, all you have to do is find the Zero's of it without a k... then there will be like 4-5 zero's and what you do is look for the one that is lowest on the y axis, and add your variable (k) to be a value just below that number. so it will only touch that point.

Answer 3

Because the polynomial is odd (largest power is odd), we know f(-inf) = -inf, and f(inf) = inf.

So we know that no matter what, it crosses the x-axis at least once.  In order to have it cross exactly once, just set k to a very large positive or negative number that will make the lower terms of the polynomial insignificant.

One solution is k=10^99

Answer 4

has one zero means that the there is only x for wich f(x) = 0.

so basiccaly we have to solve f(x)= 0.

now, what i don't understand is the formula for f(x).

if it's f(x) = (1/5) * ( x^5 - 20x^3 + 64x) + k

then we have 0 = (1/5) * ( x^5 - 20x^3 + 64x) + k. multiply by 5 =>

0 = x^5 - 20x^3 + 64x + 5k

-5k = x^5 - 20x^3 + 64x

-5k = x( x^4 - 20 X^2 + 64).

lets take  x^4 - 20 X^2 + 64.

lets say X^2 = y then  x^4 - 20 X^2 + 64  is y^2 - 20y + 64.

if we try to solve the new equation, with y, using the formula delta = b^2 - 4ac we find that in this case, delta =  144= 12^2.

which means we have two solutions for y.

the first is y1 = (20 - 12)/2 = 8/2=4  and the second is y2= (20+12)/2 = 32/2=16.

now, fisrt we have x^2 = 4 which meand x can be 2 or (-2). so we have x1 = -2 ; x2= 2 .

and y2=16=4^2 which means we have x3 = 4 and x4 = -4 .

now we go back to the equation we started from, and replace x with each and everyone of these 4 solution: -4. -2, 2 and 4.

fif x = -4 then f(x) = (1/5) * ( (-4)^5 - 20(-4)^3 + 64(-4)) + k = (1/5) * ( - 1024 + 20*64 - 64*4)+k = (1/5) * (-1024 +1280 - 256) +k= (1/5)*0 +k=K

so for f(x) to be 0 , it means that k = 0.

you can calculate for the other 3 cases.

or you could calculate directly into the formula -5k = x( x^4 - 20 X^2 + 64), if its easier for you.

remember that a negative number raised to an odd power returns a negative number.

!!!! x^ (2k+1) is negative if x is negative !!!