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# Tutor profile: Malida H.

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Malida H.
Tutor for 2 Years
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## Questions

### Subject:Pre-Calculus

TutorMe
Question:

Suppose $$\begin{equation}\lim_{x\to 2} h(x)=4\end{equation}$$ and $$\begin{equation}\lim_{x\to 2} g(x)=16\end{equation}$$ Find $$\begin{equation}\lim_{x\to 2} f(x)\end{equation}$$: 1. $$f(x)= \frac{h(x)}{g(x)}$$ 2. $$f(x)= \frac{g(x)}{h(x)}$$ 3. $$f(x)= (\sqrt{h(x)})(\sqrt{g(x)})$$

Inactive
Malida H.

Our limit rules for taking the limit of a rational expression (taking the limit of two functions and dividing them) requires that 1. both functions evaluate to a limit and are not infinite 2. the denominator is nonzero For taking the product or an exponential (a square root is simply $$x^\frac{1}{2}$$) We need that both functions evaluate to a limit and are not infinite. Luckily, we are given two functions whose limits do evaluate to a finite, nonzero number. Therefore, we can take the limit of both sides of the equation and evaluate each separately. 1. $$f(x)= \frac{h(x)}{g(x)}$$ $$\begin{equation}\lim_{x\to 2} f(x)\end{equation}$$ $$= \begin{equation}\lim_{x\to 2} \frac{h(x)}{g(x)}\end{equation}$$ $$=\frac{ \begin{equation}\lim_{x\to 2} h(x)\end{equation} } {\begin{equation}\lim_{x\to 2} g(x)\end{equation}}$$ $$= \frac {4}{16}=\frac{1}{4}$$ 2. $$f(x)= \frac{g(x)}{h(x)}$$ $$\begin{equation}\lim_{x\to 2} f(x)\end{equation}$$ $$= \begin{equation}\lim_{x\to 2} \frac{g(x)}{h(x)}\end{equation}$$ $$=\frac{ \begin{equation}\lim_{x\to 2} g(x)\end{equation} } {\begin{equation}\lim_{x\to 2} h(x)\end{equation}}$$ $$= \frac {16}{4}=4$$ 3. $$f(x)= (\sqrt{h(x)})(\sqrt{g(x)})$$ $$\begin{equation}\lim_{x\to 2} f(x)\end{equation}$$ $$=\begin{equation}\lim_{x\to 2} (\sqrt{h(x)})(\sqrt{g(x)})\end{equation}$$ $$=\begin{equation}\lim_{x\to 2} (\sqrt{h(x)})\end{equation}$$$$\begin{equation}\lim_{x\to 2} (\sqrt{g(x)})\end{equation}$$$$= (\sqrt{4})(\sqrt{16})=4(2)=8$$

### Subject:Calculus

TutorMe
Question:

Find the equation of the line tangent to $$f(x)=2x^2+5$$ at $$x=5$$.

Inactive
Malida H.

We know that a derivative of a function evaluated at any x gives us the slope of the tangent line at that specific x. So let's find the derivative of $$f(x)$$. $$\frac{d}{dx}f(x)=\frac{d}{dx}(2x^2+5)$$ Using our rules for taking derviatives, we find that $$f'(x)=4x$$ Now we must evaluate $$f'(x)$$ at $$x=5$$ So $$f'(5)=4(5)=20$$ So our slope is $$m=20$$ Let's now find the actual point where $$x=5$$. We plug in $$x=5$$ into our $$f(x)$$ $$f(5) = 2(5)^2+5=2(25)+5=50+5=55$$ So the point our tangent line passes through is $$(5,55)$$. From here, we use our slope-intercept form for the equation of a line: $$y=mx+b$$ We have: $$m=20$$ $$x=5$$ $$y=55$$ So we need to solve for b. Simply substitute these values in. $$55=20(5)+b$$ Simplifying, $$55=100+b$$ We subtract both sides by 100 to get $$b=-45$$. And now the equation for the line tangent to $$f(x)=2x^2+5$$ at $$x=5$$ is: $$y=20x-45$$

### Subject:Algebra

TutorMe
Question:

Find the equation of the line passing through the points: $$(-3, 0)$$ and $$(3, 4)$$ and give your answer in slope-intercept form.

Inactive
Malida H.

The format for slope-intercept form is $$y = m x + b$$ where $$x$$ and $$y$$ are any points on the line (arbitrary) m is slope and b is the y-intercept First, let's find our slope. Our equation for slope is $$m= \frac {y_2 -y_1}{x_2 - x_1}$$ $$x_2 = 3$$ $$x_1 = -3$$ $$y_2 = 4$$ $$y_1 = 0$$ Plugging these values in, we get $$m = \frac {4-0}{3-(-3)} = \frac {4}{6} = \frac {2}{3}$$ Now let's solve for b, our y-intercept. We can choose either point $$(-3,0)$$ or $$(3,4)$$. Let's go with $$(-3,0)$$. Right now, our equation looks like this: $$y= \frac{2}{3}x +b$$. We know: $$x=-3$$ and $$y=0$$ for this case. So now we substitute that into our equation. We get: $$0=(\frac{2}{3})(3)+b$$ And simplifying, we get $$0=2+b$$ Subtract both sides of the equation by 2 in order to solve for b. We get that $$b=-2$$ Our last step now is to substitute b back into our original equation: $$y= \frac{2}{3}x-2$$

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