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Matthew G.
High School Math Teacher 7+ years
Tutor Satisfaction Guarantee
Pre-Calculus
TutorMe
Question:

Consider the function $$f(x)=\frac{\sqrt{x}-3}{x-9}$$. Find $$\lim_{x\to9}f(x)$$.

Matthew G.
Answer:

We cannot evaluate the limit right away with direct substitution since that would lead to division by zero. Instead we will need to simplify $$f(x)$$ by using the conjugate of the numerator. The conjugate of the numerator is $$\sqrt{x}+3$$ $$\frac{\sqrt{x}-3}{x-9}*\frac{\sqrt{x}+3}{\sqrt{x}+3}$$ Which will simplify to $$f(x)=\frac{x-9}{(x-9)(\sqrt{x}+3)}$$ The binomial $$x-9$$ in the numerator and denominator will divide out to one leaving $$\frac{1}{\sqrt{x}+3}$$ We can now take the limit by using direct substitution: $$\lim_{x \to 9}\frac{1}{\sqrt{x}+3}=\frac{1}{\sqrt{9}+3}=\frac{1}{6}$$ Below is a complete answer from start to finish: $$\lim_{x\to9}\frac{\sqrt{x}-3}{x-9}=\lim_{x\to9}\frac{\sqrt{x}-3}{x-9}*\frac{\sqrt{x}+3}{\sqrt{x}+3}=\lim_{x\to9}\frac{1}{\sqrt{x}+3}=\frac{1}{\sqrt{9}+3}=\frac{1}{6}\blacksquare$$

Trigonometry
TutorMe
Question:

Verify the following: $$\frac{1-tan^2(x)}{sec^2(x)} = cos^2 (x) - sin^2 (x)$$

Matthew G.
Answer:

First decompose the expression $$\frac{1-tan^2(x)}{sec^2(x)}$$ into sines and cosines and you will get $$\frac{1-(\frac{sin(x)}{cos(x)})^2}{(\frac{1}{cos(x)})^2}$$ Then simplify the complex fraction to $$cos^2(x)-sin^2(x)$$. $$\blacksquare$$

Algebra
TutorMe
Question:

Completely factor $$f(x)=2x^4+7x^3-4x^2-27x-12$$ and find all the zeros of the polynomial.

Matthew G.
Answer:

By rational roots theorem we get the following possible zeros of the polynomial: $$\frac{+}{-}1,2,3,6,9,18, \frac{1}{2},\frac{3}{2},\frac{9}{2}$$ By testing and using synthetic division you can find 2 as a root and leaves $$2x^3+11x^2+18x+9$$ as the quotient. Which means that $$f(x)$$ can be factored to $$f(x)=(x-2)(2x^3+11+18x^2+9)$$ By continuing to test zeros you will find that -3 is also a root off $$f(x)$$ which will show that $$f(x)= (x-2)(x+3)(2x^2+5x+3)$$. The quadratic factor $$2x^2+5x+3$$ can be factor into a product of linear factors $$(2x+3)(x+1)$$. The complete factorization of $$f(x)$$ is $$(x-2)(x+3)(2x+3)(x+1)$$. By using the zero product property we can find that $$f(x)$$ has the following zeros: $$x=-3,-\frac{3}{2},-1,$$ and $$2$$.

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