# Tutor profile: Justin L.

## Questions

### Subject: Pre-Calculus

How many real solutions does $$x^2-2x + 1 = 0$$ have?

One approach is to factor the equation $$\begin{align*} 0 &=x^2-2x+1 \\ &= ( x - 1)^2 \end{align*}$$ Since the only solution is $$x=1$$, there is 1 real solution. *NOTE: Although the root at $$x=1$$ has multiplicity 2, the value 1 is the only root. A second approach is to look at the discriminant $$D$$. Using the form $$ax^2 + bx + c = 0$$ we can substitute $$a=1, b=-2,$$ and $$c=1$$, to evaluate $$D=b^2-4ac$$ and look at whether $$D$$ is 0, positive, or negative. When substituting, we see $$D = 4 - 4(1)(1) = 0$$. Since the discriminant is $$0$$ there is 1 real solution. This confirms our findings.

### Subject: Calculus

For the following function $$f$$, determine if it is continuous on $$R$$. If so find its derivative. $$f(x) = e^x\sin(x)$$

Since $$f$$ can be written as $$f(x) = g(x)\cdot h(x)$$ where $$g(x)=e^x$$ and $$h(x) = \sin(x)$$ and both $$g$$ and $$h$$ are continuous on $$R$$, then $$f$$ is continuous on $$R$$. Continuing with our definitions of $$g$$ and $$h$$, we can use the Product Rule on $$f$$. $$\begin{align*} f'(x) &= g'(x)\cdot h(x) + g(x)\cdot h'(x) \\ &= e(x) \cdot \sin(x) + e^x\cdot \cos(x) \\ &= e^x\cdot[s\sin(x) + \cos(x)] \end{align*}$$

### Subject: Algebra

A farmer wants to build a rectangular pen that is twice as long as it is wide. He has 300 feet of fencing material. What should the width of the pen be?

First, we let $$W$$ denote the width. This is what we would like to find. Now we write our list of facts: $$(1)] P=300$$ where $$P$$ is the perimeter (the amount of fencing material) $$(2)] P=2W + 2L$$ since the pen is a rectangle $$(3)] L = 2W$$ since the farmer wants the length to be twice the width Since we know the value of $$P$$ from $$(1)$$, we can use this in $$(2)$$. This gives us another fact: $$300 = 2W + 2L$$ Now we can use $$(3)$$ with our new fact $$300=2W+2L$$ to say $$300 = 2W + 2(2W)$$ where we have substituted for $$L$$. We combine like terms and have $$300 = 6W$$ Therefore $$W=50$$ Since we were looking for $$W$$ this problem is done. The width should be 50 feet.

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