Tutor profile: Pearce R.

Calculate the $$\frac{dy}{dx}$$ where $(y = \frac{x+3}{x^2-2}$)

We can split the expression into $$y = (x+3)(x^2-2)^{-1}$$ and use the product rule and the chain rule. $(\frac{dy}{dx} = 1*(x^2-2)^{-1}+ (x+3)(-(x^2-2)^{-2})(2x)$) $(\frac{dy}{dx} = \frac{1}{(x^2-2)}- \frac{2x^2+6x}{(x^2-2)^{2}}$) $(\frac{dy}{dx} = \frac{-x^2-6x-2}{(x^2-2)^{2}}$)

What is the value of x after the following code executes? int x = 0 for(int i = 0; i < 10; i++){ x = x+i }

This is an example of a for loop. x is initialized to 0, as is i. The first time, x=x+i evaluates to x = 0+0, so x remains 0. Then, i is incremented to 1 (by i++ at the end of the for loop) Now, x=x+i evaluates to x = 0+1, so x=1 The loop continues to add i to x, x will grow with the pattern 0, 1, 3, 6, 10, 15, ... When i reaches the value 9, x=x+i executes one final time, after which i is incremented to 10 and the loop executes. Thus, x will take values 0,1,3,6,10,15,21,28,36,45 The code will end with x having a final value of 45

Find the roots of the equation ($y = 6x^2 +12x+6$ )

The roots of the equation are the points where the function drawn on the XY plane crosses the x axis, in other words where y=0, what are the value(s) of x? First, we plug in 0 for y, which gives us $(0 = 6x^2+12x+6$) Then, because $$\frac{0}{6} = 0$$, we can write the equation as $(0 = x^2+2x+1$) Next, we must factor the expression, which can be factored as $(0 = (x+1)(x+1)$) Which is clearly true only when $(x=-1$) Another method of solving it is to use the quadratic equation, if we rewrite the equation as the form $(y=ax^2+bx+c$)Thus $$a=6$$, $$b=12$$, $$c=6$$ So by the quadratic equation the roots are $(x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$) or $(x=\frac{-12 \pm \sqrt{12^2-4*6*6}}{2*6} = \frac{-12 \pm \sqrt{0}}{12} = -1$)