# Tutor profile: Brandon G.

## Questions

### Subject: Pre-Calculus

The function $$f(x) = k(9-x-x^3)$$ is one-to-one and $$f^{-1}(77) = -6.$$ Find $$k.$$

By definition of an inverse function, if $$f^{-1}(y) = x$$ then we must have $$f(x) = y.$$ This is the case for a unique value $$y$$ given that $$f$$ is one-to-one. Hence, we must have: $$\begin{align*} f(-6) = 77\\ k(9-(-6)-(-6)^3) = 77 \end{align*}$$ Solving this last expression will give us $$231k = 77.$$ Solving for $$k$$, we have $$k= \frac{77}{231} = \frac{1}{3}.$$

### Subject: Calculus

Differentiate $$F(x) = x^7 \ln(3x).$$

We must use the product rule: $$\begin{align} \frac{d}{dy}\left(f\cdot g \right)(x) = f'(g(x)) + f(g'(x)) \end{align}$$ In this case, we have $$f= x^7$$ and $$g = \ln 3x.$$ This will give us: $$\begin{align*} F'(x) &= 7x^{6} \ln 3x + x^7 \cdot \frac{1}{x}\\[5pt] &= 7x^6 \ln 3x + x^6 \end{align*}$$ This last expression is also equivalent to $$x^6(7\ln (3x) + 1).$$ $$F(x)= \frac{e^{3x}}{x^9}$$ For this problem, we can also use the product rule noting that the function is equivalent to $$e^{3x}\cdot x^{-9}.$$ We will have: $$\begin{align*} F'(x) &= 3e^{3x}x^{-9} + e^{3x}(-9x^{-10})\\[5pt] &= \frac{3e^{3x}}{x^{9}} - \frac{9e^{3x}}{x^{10}}\\[5pt] &= \frac{3xe^{3x}-9e^{3x}}{x^{10}}\\[5pt] &= \frac{3e^{3x}(x-3)}{x^{10}} \end{align*}$$

### Subject: Statistics

Assume that the distribution of SAT math scores has a mean $$\mu = 600$$ and a standard deviation of $$\sigma = 100.$$ Suppose that a simple random sample of $$n = 25$$ students took the test and got an average of $$\bar{x} = 700$$ with a standard deviation of $$s = 90.$$ Is it correct to conclude that this group of students did better than the average? Use 0.05 significance.

The hypotheses are: $$H_0: \mu = 600$$ $$H_1: \mu > 600$$ And hence this is a one-tailed $$z$$-test for the mean (as we have information regarding the population's standard deviation). We will calculate the $$z$$-score as follows: $$z = \frac{\bar{x}-\mu}{\sigma/\sqrt{n}}$$ To obtain $$z = 5.$$ Then, we compute the $$p$$-value using the normal distribution: $$P(z > 5) \approx 0.0001$$. Since this is smaller than 0.05, we reject the null hypothesis. We conclude that the group of students did better than average. Another way to run this test is to use the upper critical value of $$z_c = 1.645.$$ Since our test statistic is greater than this, we reject the null hypothesis.

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