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Justin L.
Mathematician & Software Developer
Tutor Satisfaction Guarantee
Pre-Calculus
TutorMe
Question:

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

Justin L.

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.

Calculus
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Question:

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

Justin L.

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*}

Algebra
TutorMe
Question:

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?

Justin L.

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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