A question that has me stumped?

Question

this question is in two parts, and I'm finding it quite difficult to get my head around this as the study material I have doesn't really explain things too well.

a) Examine the behaviour of the function f(x) = x^4 - 4x^2 - 12, include in your answer the exact values of the vertical and horizontal intercepts, stationary points, any asyptotes, and a comment on the functions continuity.

b) From the function above, create this function with a restricted domain such that the function now has an inverse that is a function

Any help with this would be greatly appreciated !!!

Answer 1

f(x) = x^4 - 4x^2 - 12

To solve for the stationary points (also known as local extrema), we find the first derivative and then make it 0.

f'(x) = 4x^3 - 8x

f'(x) = 4x(x^2 - 2)

Making f'(x) equal to 0,

0 = 4x(x^2 - 2)

Divide both sides by 4,

0 = x(x^2 - 2)

Now, we solve this by factoring.

0 = x[x - sqrt(2)] [x + sqrt(2)]

This subsequently leads to the solutions x = {0, sqrt(2), -sqrt(2)}

These are our critical numbers. To determine intervals of increase and decrease, we need to test a single point between each critical number. If that point we test is positive, then we have an increasing function on that interval. If that point we test is negative, then we have a decreasing function on that interval.

Testing x = -10 for f'(x) = 4x(x^2 - 2) gives us a negative solution.
Therefore, f(x) is decreasing on (-infinity, -sqrt(2)]

Testing x = -1 gives us a positive solution, x = 1 a negative solution, and x = 10 a positive solution.

Therefore

f(x) is decreasing on (-infinity, -sqrt(2)] U [0, sqrt(2)]
f(x) is increasing on [-sqrt(2), 0] U [sqrt(2), infinity)

This means we have a local minimum at x = -sqrt(2) and
x = sqrt(2).

We have a local maximum at x = 0 and x = sqrt(2)

So our stationary points are at
(-sqrt(2) , f(-sqrt(2)) ) , (sqrt(2) , f(sqrt(2)) ) , (0, f(0))

Since this is a polynomial, there aren't any vertical or horizontal asymptotes.

Answer 2

a) First, find the intercepts. This means find the value of f(0), along with the values of f(x) such that f(x) = 0. Finding f(0) is easy, because when you plug in zero you get -12. So that's where it crosses the f(x) axis. To find the values that give f(x) = 0, just set it equal to zero and sove for x. This factors out nicely:
0 = x^4 - 4x^2 - 12; 0 = (x^2 + 2)(x^2 - 6). So x = √6 and -√6 are the only real values. Those are the four places where it crosses the x axis.

To find stationary points, find when f'(x) = 0. So just take the derivative of f(x), set it equal to zero, and solve for x. As for asymptopes or discontinuity, I don't see any reason to suspect either.

b) Remember a function is by definition a process of taking an x value and assigning it to one and only one y value. Our current function has a graph where more than one x value has the same f(x) value. This is OK, because it's still a function. However, it would mean the inverse would have multiple y values for the same x value. So pick a range of x where not only every x value has a unique y value, but that none of the y values repeat. If you take a look at the graph when x > √6, this should work.

Answer 3

To find the vertical intercepts, set x = 0, solve for y.

(0,-12) is the one and only vertical intercept.

To find the horizontal intercepts, set y = 0, solve for x. This requires factoring the polynomial. A fourth degree equation will have four solutions, but these solutions may be doubled.

Stationary points. Take the derivative, set that equal to zero. Solve for x. Where x = 0 is a stationary point.

Asymptotes. You should be able to look at this equation and tell that there are no asymptotes.

Continuity. Same thing, you should be able to look at this and tell it is continuous for the entire domain.

b) Restrict the domain in such a way that a horizontal line would not cross the curve twice. (between two stationary points, for instance).

Answer 4

asymotope at -12 because x tends to 0 and so (x^4 - 4x^2) tends to 0 therfore there is an asymatope at y= -12 (on the y axis if y = f(x))

Points of intercept: sub in x = 0 and y = 0 (assuming y =f(x))

y=(0)^4 - 4(0)^2 -12
y=-12 point of intercept (vertical) (0,12)

0=X^4 - 4X^2 -12
0=(x^2 + 2)(X^2 - 6)

Therfore X^2 = -2 Therfore X = root -2 (doesnt exsit unless you call it 2i imaginary numbers crazy shizzle)

Therfore X^2 = 6 Therfore X = + or - root 6

Points of intercept (horizontal) at (+ root 6,0) and at (- root 6,0)

stationary points (differentiate the equation to find the equation of the graident of the curve)

DY/DX = 4x^3 - 8X

Make M (gradient) = 0 which gives a stationary point where
0=4x^3 - 8X

solve cant b botherd sorry

sub x values into f(X) equation to get y values which gives you your co ordinates

b)
inverse f(X)^-1

x= Y^4 - 4Y^2 - 12

rearange to get y = this is the inverse function

hoped that helped if it did at all i h8 maths 2 a point this is that point.