# Tutor profile: Robert B.

## Questions

### Subject: SAT

Simplify the expression $$\frac{18x^3 + 9x^2 - 36x}{9x^2}.$$

We may write the expression in the equivalent form $$\frac{1}{9x^2}(18x^3 + 9x^2 - 36x),$$ and then use the distributive property to find that it is equal to $$\frac{18x^3}{9x^2}+\frac{9x^2}{9x^2} - \frac{36x}{9x^2}.$$ By reducing the fractions, and applying the rule $$\frac{x^b}{x^a} = x^{b-a},$$ the expression becomes $$2x^{3-2} + 1x^{2-2} - 4x^{1-2} = 2x + 1 -4x^{-1}.$$

### Subject: Calculus

Consider the polynomial function $$y(x) = x^2 + x + 1.$$ Compute the corresponding derivative $$y'$$ and determine where $$y$$ is increasing.

We recall the power rule for derivatives: if $$f(x) = x^{n}$$ for some natural number $$n > 0,$$ then $$f'(x) = nx^{n-1}.$$ We also recall the sum rule for derivatives: if $$f, g, h$$ are functions then $$\frac{d}{dx} (f(x) + g(x) + h(x)) = \frac{d}{dx} f(x) + \frac{d}{dx} g(x) + \frac{d}{dx} h(x)$$. In our particular case, $$f(x) = x^2,$$ (so $$f'(x) = 2x$$), $$g(x) = x$$ (so $$g'(x) = 1 $$), and $$h(x) = 1$$ (so $$h'(x) = 0$$). Therefore, by the Sum Rule, $$y'(x) = 2x + 1.$$ For the second half of the question, note that $$y$$ is increasing if and only if $$y' > 0.$$ In particular, we must have $$y'(x) = 2x + 1 > 0.$$ By setting $$0 = y'(x) = 2x + 1,$$ we find that $$y' > 0$$ if and only if $$x > - \frac{1}{2}.$$ Therefore $$y$$ is increasing whenever $$x > - \frac{1}{2}.$$

### Subject: Algebra

Consider the quadratic function $$f(x) = x^2 + 1.$$ Deduce the number of real solutions that f has, and solve for the particular solutions.

For the first half of the question, we use the discriminant formula $$D = b^2 - 4ac.$$ In the given function, $$b =0,$$ $$a=1,$$ and $$c=1$$, so that $(D = 0 + (-4)(1)(1) = -4 < 0.$) Because the discriminant $$D$$ is less than zero, we conclude that $$f$$ has two complex solutions, i.e., zero real solutions. For the second half of the question we set $$f(x) = 0$$ so that $$0 = x^2 + 1$$. Equivalently, $$x^2 = -1,$$ and we may solve for $$x$$ (up to sign) by taking the square root of both sides of the equation. In particular, $$x = \pm \sqrt{-1} = \pm i.$$

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