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# Tutor profile: Robert B.

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Robert B.
Tutor and Educator for 5+ years
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## Questions

### Subject:Pre-Calculus

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Question:

Find $$f^{-1}(x)$$ if $$f(x)=e^{-x^2+6x-9} -\frac{7}{2}$$

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Robert B.

Let $$y=f(x)$$ $$y=e^{-x^2+6x-9} -\frac{7}{2}$$ $$y+\frac{7}{2}=e^{-x^2+6x-9}$$ $$\mathrm{ln}(y+\frac{7}{2})=-x^2+6x-9$$ $$-\mathrm{ln}(y+\frac{7}{2})=x^2-6x+9$$ $$-\mathrm{ln}(y+\frac{7}{2})=(x-3)^2$$ $$\sqrt{-\mathrm{ln}(y+\frac{7}{2})}=x-3$$ $$\sqrt{-\mathrm{ln}(y+\frac{7}{2})}+3=x$$ By definition $$f^{-1}(y)=x$$ So, $$\sqrt{-\mathrm{ln}(y+\frac{7}{2})}+3=f^{-1}(y)$$ and, since the variable is arbitrary: $$f^{-1}(x)= \sqrt{-\mathrm{ln}(x+\frac{7}{2})} +3$$

### Subject:Basic Math

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Question:

Factor: $$2x^2 + 5x - 12$$

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Robert B.

Method 1: Complete the square Rearrange the equation as follows: $$2x^2 + 5x = 12$$ Multiply the equation by 2 $$2(2x^2 + 5x = 12)$$ (*) $$4x^2 + 10x = 24$$ to complete the square on the left we note that $$4x^2=(2x)^2$$ so we will have a square of the form $$(2x + r)^2$$, which gives us: $$4x^2 +4rx + r^2 = 4x^2 + 10x + c$$ comparing like powers of $$x$$ we see that $$4rx = 10x$$ $$r = \frac{5}{2}$$ and $$c = r^2$$ $$c = \frac{25}{4}$$ Returning to our equation marked (*) $$4x^2 + 10x + \frac{25}{4} = 24 + \frac{25}{4}$$ $$(2x + \frac{5}{2})^2 = \frac{96 + 25}{4}$$ $$(2x + \frac{5}{2})^2 - \frac{121}{4} = 0$$ using the difference of squares formula we have: $$(2x + \frac{5}{2} + \frac{11}{2})(2x + \frac{5}{2} - \frac{11}{2}) = 0$$ $$(2x + \frac{16}{2})(2x - \frac{6}{2}) = 0$$ $$(2x + 8)(2x - 3) = 0$$ $$2(x + 4)(2x - 3) = 0$$ $$(x+4)(2x-3) = 0$$ so, $$2x^2 + 5x -12 = (2x-3)(x+4)$$ Method 2: Quadratic Formula The quadratic formula for a quadratic of the form: $$ax^2 + bx +c = 0$$ is: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ for our equation: $$2x^2 + 5x - 12 = 0$$ $$a = 2$$, $$b = 5$$, and $$c = -12$$ so, $$x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-12)}}{2(2)}$$ $$x = \frac{-5 \pm \sqrt{25 +96}}{4}$$ $$x = \frac{-5 \pm \sqrt{121}}{4}$$ $$x = \frac{-5 \pm 11}{4}$$ $$x_1 = \frac{6}{4}=/frac{3}{2}$$ $$x_2 = \frac{-16}{4} = -4$$ so, $$2x^2 + 5x -12 = (x - \frac{3}{2})(x + 4)$$ or $$(2x - 3)(x+4)$$

### Subject:Calculus

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Question:

Berry Creamy Ice Cream wants to create a cone with a volume of $$\frac{16\sqrt{2}\pi}{3}mL$$ while using the least amount of waffle as possible. What dimensions ($$r$$ and $$h$$) should they use for their cone?

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Robert B.

"[U]sing the least amount of waffle possible" indicates this is a minimization question; we want to minimize the surface area using the given volume. We will need the equations for the volume of a circular cone and the surface area of the curved portion of the cone (not including the area of the base, as the ice cream cone is open). $$V_{cone} = \frac{1}{3}\pi r^2 h$$ $$S_{cone}=\pi r l$$; where $$l$$ is the slant height of the cone, given by $$l= \sqrt{r^2 + h^2}$$ Begin by using the volume equation to find $$h$$ in terms of $$r$$. $$V_{cone} = \frac{1}{3}\pi r^2 h$$ $$\frac{1}{3}\pi r^2 h=\frac{16\sqrt{2}\pi}{3}$$ $$h=\frac{16\sqrt{2}}{r^2}$$ Then use this value for $$h$$ in the surface area equation. $$S_{cone}=\pi r\sqrt{r^2 + h^2}$$ $$S=\pi r\sqrt{r^2 + (\frac{16\sqrt{2}}{r^2})^2}$$ $$S=\pi r\sqrt{r^2 +\frac{512}{r^4}}$$ bringing the $$r$$ into the square root gives: $$S=\pi \sqrt{r^4 + \frac{512}{r^2}}$$ This is the expression we need to minimize. So, we will take the derivative of the surface area equation. $$\frac{\mathrm{d}S}{\mathrm{dr}}=\frac{\mathrm{d}}{\mathrm{dr}}(\pi \sqrt{r^4 + \frac{512}{r^2}})$$ $$S' = \pi \frac{4r^3 + \frac{512}{(-2)r^3}}{2\sqrt{r^4 + \frac{512}{r^2}}}$$ $$S' = \pi \frac{2r^3 - \frac{128}{r^3}}{\sqrt{r^4 + \frac{512}{r^2}}}$$ To minimize $$S$$ we need to find when $$S' = 0$$, which will occur when the numerator of the expression is $$0$$ (as the denominator is always non-zero). Thus, we need: $$2r^3 - \frac{128}{r^3} = 0$$ $$2r^3 = \frac{128}{r^3}$$ $$r^6 = 64$$ which as real-valued answers of $$r=2$$ and $$r=-2$$. Negative values of $$r$$ do not make sense for our problem, so we have determined that $$r=2$$ should minimize our surface area. As a quick check, note that for $$r=1 < 2$$ $$S' <0$$ and for $$r=4 > 2$$ $$S' > 0$$ proving $$S(2)$$ is a minimum. When $$r=2$$ we have $$h=\frac{16\sqrt{2}}{r^2}$$ $$h=\frac{16\sqrt{2}}{(2)^2}$$ $$h=\frac{16\sqrt{2}}{4}$$ $$h=4\sqrt{2}$$ The given volume has units of $$mL$$, where $$1 mL = 1 cm^3$$. Therefore, $$r$$ and $$h$$ have units of $$cm$$. So the dimensions of the cone needed to minimize the amount of waffle for the given volume are $$r=2cm$$ and $$h=4\sqrt{2}cm$$.

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