Dawson 201-015-RE: Secondary V math tutoring

The original denominator requires $x\ne3$. Factor the numerator: $x^2-9=(x-3)(x+3)$. For allowed inputs, the equation becomes $x+3=6$, giving the candidate $x=3$. But that input makes the original expression undefined. Therefore there is no solution. Cancellation simplifies the rule on its domain; it does not restore an excluded input.

The squared quantity vanishes at $x=-2$, so the vertex is $(-2,18)$. Since $(x+2)^2\ge0$, multiplying it by −2 gives a nonpositive term: $q(x)\le18$. The graph opens downward and its maximum value is 18. For intercepts, set the output to zero: $(x+2)^2=9$, so $x+2=\pm3$ and $x=-5$ or $x=1$. The intercepts are $(-5,0)$ and $(1,0)$, equally spaced about $x=-2$.

On the stated domain, $x+1\ge0$, and the outputs cover $[0,\infty)$. From $y=(x+1)^2$, take the nonnegative square root: $x=\sqrt y-1$. Thus $f^{-1}(x)=\sqrt x-1$, with domain $[0,\infty)$ and range $[-1,\infty)$. In particular, $f^{-1}(9)=2$, and $f(2)=9$. Without the restriction, distinct inputs such as 0 and −2 give the same output; the full quadratic would not have an inverse function. The inverse reverses the rule; it is not $1/f(x)$.

Both logarithm arguments must be positive: $x-1\gt0$ and $x+1\gt0$, so $x\gt1$. On that domain the sum is $\log_2((x-1)(x+1))$. Rewrite the logarithmic equation in exponential form: $x^2-1=2^3=8$. The algebraic candidates are $x=\pm3$; only $x=3$ is in the original domain. Check: $\log_2 2+\log_2 4=1+2=3$. At −3 the original logarithms are not real.

Pythagoras gives the adjacent side as $\sqrt{25^2-7^2}=\sqrt{576}=24$ cm; take the positive root for a length. The known opposite side and hypotenuse give $\sin\theta=7/25$, so $\theta=\arcsin(7/25)\approx16.3^\circ$. Use degree mode. A second check is $\tan\theta=7/24$. The angle is acute and the shorter leg is opposite the smaller acute angle. Using $7/24$ as a sine ratio would confuse the adjacent side with the hypotenuse.