Maths Topic 5: Real Functions Trivia Quiz

Answer: -1

To solve this question, we need to substitute x=-2, for each x term in the function.
So,
If f(x)=x^2+3x+1
f(-2)=(-2)^2+3(-2)+1
f(-2)=4-6+1
f(-2)=-1

Answer: 234

To solve this question, we need to substitute x=3, for each x term in the function
So,
If f(x)=x^5-x^2
f(3)=(3)^5-(3)^2
f(3)=243-9
f(3)=234

Answer: f(-x)=-f(x)

To show that a function is odd, it must first be proved that it isn't even by using the rule f(x)=f(-x). Then using f(-x)=-f(x), it can be proven whether or not a function is odd.

Answer: f(x)=x^2+3

For a function to be even, it must first satisfy the rule f(x)=f(-x). Out of the possible answers, f(x)=x^2+3 is the only possible solution.
So,
f(x)=x^2+3
f(-x)=(-x)^2+3
f(-x)=x^2+3
f(-x)=f(x)
Therefore, f(x)=x^2+3 is the correct solution.

Answer: x>5

If x=5 was subbed into the equation, then the denominator would equal 0, giving an undefined result. If any number less than 5 was subbed in, then this would give a non-real result. Therefore, any number greater than 5 would give a real result, making the domain x greater than 5.

Answer: all real y, y can't equal 0

The function f(x)=1/(x-4) resembles a hyperbola with a horizontal asymptote at y=0. Therefore, the range of f(x) is all real y, where y can't be equal 0.

Answer: y=1.5

To find the y-intercept of a function, we simply need to sub in x=0.
So, subbing in x=0 gives,
2y=3
y=3/2
y=1.5

Answer: x=-3, x=1

To find the x-intercepts, we simply sub in y=0.
This gives, x^2+2x-3=0, which we can solve as a quadratic equation.
So,
x^2+2x-3=0
(x+3)(x-1)=0
Therefore, x=-3, x=1

Answer: (3,1)

The general equation of a circle is (x-h)^2+(y-k)^2=r^2, where (h,k) is the centre of the circle and r is the radius. From the information in the question, we can work out that the centre of the circle is (3,1).

Answer: y=sqrt(9-x^2)

The general equation for a semi-circle that lies above the x-axis is y=sqrt(r^2-x^2), where r is the radius of the semi-circle. The question tell us that the radius is 3, therefore, we can work out that the correct equation for the semi-circle is y=sqrt(9-x^2).

Answer: y=-1

To solve this question, we would first need to use the graph y=abs(x). From this, we can see that the smallest y-value lies at y=0. Using abs(x)-1, this pushes abs(x) down 1 unit, making the smallest y-value at y=-1.

Answer: y=1

To solve this question, we can simply sub in x=0. This gives us y=a^0. Now using index laws, any number to the power of 0 will give a result of 1.
Therefore, the y-intercept of y=a^x and y=a^(-x) will always be equal to 1.

Answer: 1/4

To solve this question, we first need to simplify the given equation.
So,
(x-2)/(x^2-4)
=(x-2)/((x-2)(x+2))
=1/(x+2)
Now sub in x=2
=1/(2+2)
=1/4
Therefore, the limit is 1/4.

Answer: x=-2, y=0

To solve this question, we need to find out what values x and y cannot be equal to.
For the vertical asymptote,
x+2=0
x=-2
Therefore, the vertical asymptote is at x=-2, since if we sub this in, we would get an undefined result.
For the horizontal asymptote, since the graph isn't translated up or down, and it asymptotes towards the x-axis, this makes the horizontal asymptote y=0.

Answer: x>3

In the graph f(x)=logx, there is a vertical asymptote at x=0, making the domain x greater than 0. Therefore, for log(x-3), the graph is shifted 3 units to the right. The makes the vertical asymptote at x=3, and the domain x greater than 3.