# Tutor profile: Kapil G.

## Questions

### Subject: Pre-Calculus

Find the domain of the function 1/(x +5).

In order to find the domain of the given rational function 1/x+5, we should remember that the domain of a rational function is all real numbers or (-infinity, infinity) except the number which makes the denominator zero. To find the number which makes the denominator zero, we set the denominator equal to 0 and solve the equation. Here, the denominator is x + 5 so we will set x + 5 equal to 0 and solve for x. x + 5 = 0 (we will subtract 5 from both sides to get x by itself) x + 5 - 5 = 0 - 5 x = -5 Hence, the domain of the given rational function would be all real numbers except -5. We can also write it in interval notation as follows: (-infinity, -5) U (-5, infinity) We have written -5 with an open circle because an open circle indicates that we have excluded the value.

### Subject: Calculus

The side of a square is increasing at the rate of 2 meters per second. Find the rate at which the area of the square increasing when the side of the square measured 4 meters.

Let the side of the square be s meters. Now it says that the rate of change of the side is 2 meters per second. Hence, ds/dt = 2 m/s. Now the area of the square formula is A = side^2, here side = s so it would be A = s^2. To find the rate of change of area we will derive both sides with respect to t: dA/dt = 2s * ds/dt ( To find the derivative of s^2, we will use the power rule of derivative. As per the power rule the derivative of x^n is nx^(n-1) so if we apply this to s^2 it would be 2s^(2-1) or 2s^1 or just 2s. Then as per the chain rule we will derive s as well so we will multiply it by ds/dt) Now we know to find the rate of change of area, dA/dt, when the side of the square is 4 meters or s = 4. So we will plug s = 4 and ds/dt = 2 into dA/dt = 2s * ds/dt. dA/dt = 2(4)(2) dA/dt = 16 Hence, the rate of change of the area would be 16 square meters per second.

### Subject: Algebra

Solve for x: √(x-3) = -4

We have to solve the given square root equation for x. First, we will get rid of the square root from the left side by squaring both sides: [√(x-3)]^2 = (-4)^2 The square and square root will cancel each other on the left side and we would be left with just x - 3 on the left side. On the right side, (-4)^2 would be -4 times -4 = 16 so the equation would become x - 3 = 16. Then to get x by itself we will add 3 both sides. So it would look like x - 3 + 3 = 16 + 3. Then -3 and 3 will cancel each other on the left side and we would be left with x = 19. Now we will plug x = 19 into the original equation to check whether it satisfies the original equation or not. √(19 - 3) = -4 √16 = -4 4 = -4 As shown above, if we plug x = 19 into the original equation the left side would be the square root of 16 which is 4. We know that 4 is not equal to -4. Hence, x = 19 is not its solution. So for this equation there is "No Solution".

## Contact tutor

needs and Kapil will reply soon.