Functions & Curve Sketching

Sketching the curve and finding asymptotes.

Begin by rewriting the function in a more revealing form. Perform polynomial long division on $\frac{2x(x^2-5)}{x^2-4}$:$$2x(x^2-5)=2x(x^2-4)-2x⟹\frac{2x(x^2-5)}{x^2-4}=2x-\frac{2x}{(x-2)(x+2)}.$$This decomposition makes the structure transparent. As $|x|\to \infty$, the fraction $\frac{2x}{(x-2)(x+2)}\to 0$, so the oblique asymptote is $y=2x$. The vertical asymptotes occur where the denominator vanishes: $x=2$ and $x=-2$.

To find the tangent at the origin, differentiate:$$\frac{dy}{dx}=2-\frac{2(x^2-4)-2x\cdot 2x}{(x^2-4{)}^2}=2-\frac{-2x^2-8}{(x^2-4{)}^2}.$$At $x=0$: $\frac{dy}{dx}{|}_{x=0}=2-\frac{-8}{16}=2+\frac{1}{2}=\frac{5}{2}$.

Since $y(0)=0$, the tangent at the origin is $y=\frac{5x}{2}$.

The curve passes through the origin, has zeros at $x=\pm \sqrt{5}$, and approaches $y=2x$ from above or below depending on the branch.

Part (i). Rearrange $3x(x^2-5)=(x^2-4)(x+3)$ as follows (assuming $x\ne \pm 2$):$$\frac{2x(x^2-5)}{x^2-4}=\frac{2}{3}x+2.$$Count the intersections of $y=\frac{2}{3}x+2$ with the sketched curve. This line has gradient $\frac{2}{3}$, which is shallower than the oblique asymptote $y=2x$, and crosses the $y$-axis at $2$. A careful reading of the sketch shows the line cuts the curve in three points, so there are three real roots.

Part (ii). Rearrange $4x(x^2-5)=(x^2-4)(5x-2)$ (assuming $x\ne \pm 2$):$$\frac{2x(x^2-5)}{x^2-4}=\frac{5x}{2}-1.$$The line $y=\frac{5x}{2}-1$ is parallel to the tangent at the origin $y=\frac{5x}{2}$ but shifted one unit downwards. Since $y=\frac{5x}{2}$ is tangent to the curve at the origin, a parallel line just below it touches the curve at one fewer point there and does not cross the curve near the origin. Checking the other branches confirms the line meets the curve at exactly one point, giving one real root.

Part (iii). Factor $4x^2(x^2-5{)}^2=(x^2-4{)}^2(x^2+1)$ as (assuming $x\ne \pm 2$):$$\frac{2x(x^2-5)}{x^2-4}=\pm \sqrt{x^2+1}.$$The curves $y=\sqrt{x^2+1}$ and $y=-\sqrt{x^2+1}$ are the two branches of a hyperbola with asymptotes $y=|x|$ and $y=-|x|$ respectively, since $\sqrt{x^2+1}\approx |x|$ for large $|x|$ and $\sqrt{x^2+1}\ge 1$ always. Each branch has asymptotes parallel to the oblique asymptote of the original curve, and each intersects the curve in three points. Combining both branches yields six real roots in total.

Part (i).

The function $y=\sqrt{1-x}+\sqrt{3+x}$ requires both square roots to be defined, so we need $1-x\ge 0$ and $3+x\ge 0$, giving the domain $-3\le x\le 1$. At both endpoints the curve takes the value $y=2$: at $x=1$ we get $\sqrt{0}+\sqrt{4}=2$, and at $x=-3$ we get $\sqrt{4}+\sqrt{0}=2$.

The function is symmetric about $x=-1$: replacing $x$ by $-2-x$ swaps the two square roots, leaving $y$ unchanged. The maximum therefore occurs at $x=-1$, giving $$y_{\max }=\sqrt{2}+\sqrt{2}=2\sqrt{2}\approx \mathrm{2.83.}$$ Since $2\sqrt{2}>2$, the curve rises above the endpoints — it is a smooth arch between $(-3,2)$ and $(1,2)$, symmetric about $x=-1$, with the gradient becoming infinite at both endpoints (each term behaves like a square-root near its zero).

Now sketch $y=x+1$ on the same axes. This straight line passes through $(-1,0)$ and $(1,2)$. The endpoint $(1,2)$ lies on both curves simultaneously, so $x=1$ is a solution. For $x<1$ the arch lies strictly above the line (the arch value at $x=-1$ is $2\sqrt{2}\approx 2.83$, while the line gives $0$), so the two graphs meet only at $x=1$. Hence $x=\boxed{1}$ is the unique solution.

Part (ii).

Rewrite the equation as $$2\sqrt{1-x}=\sqrt{3+x}+\sqrt{3-x}.$$ The right-hand side requires $3-x\ge 0$ as well, so the domain is now $-3\le x\le 1$. On this domain, $y_{\text{RHS}}=\sqrt{3+x}+\sqrt{3-x}$ is symmetric about $x=0$ with maximum $2\sqrt{3}$ at $x=0$, and takes value $\sqrt{6}$ at both $x=\pm 3$. The left-hand side $y_{\text{LHS}}=2\sqrt{1-x}$ is a decreasing function, equal to $2\sqrt{4}=4$ at $x=-3$ and $0$ at $x=1$. A sketch shows the two curves cross exactly once, at a negative value of $x$.

To find it algebraically, square both sides: $$4(1-x)=(3+x)+2\sqrt{(3+x)(3-x)}+(3-x)=6+2\sqrt{9-x^2}.$$ Rearrange: $$4-4x-6=2\sqrt{9-x^2}⟹-2-4x=2\sqrt{9-x^2}⟹-1-2x=\sqrt{9-x^2}.$$ For the square root to be non-negative we need $-1-2x\ge 0$, i.e. $x\le -\frac{1}{2}$. Square again: $$(1+2x{)}^2=9-x^2⟹1+4x+4x^2=9-x^2⟹5x^2+4x-8=0.$$ The roots are $$x=\frac{-4\pm \sqrt{16+160}}{10}=\frac{-4\pm \sqrt{176}}{10}=\frac{-2\pm 2\sqrt{11}}{5}.$$ Since $\sqrt{11}\approx 3.32$, the two roots are approximately $x\approx 0.92$ and $x\approx -1.72$. Only $x=\frac{-2-2\sqrt{11}}{5}$ satisfies $x\le -\frac{1}{2}$, so $$x=\frac{-2-2\sqrt{11}}{5}.$$

Part (i) — Tangent from the origin. For a line $y=mx$ to be tangent to $y=\ln x$, two conditions must be satisfied simultaneously at the point of contact $(x_0,\ln x_0)$: firstly the point lies on the line, so $\ln x_0=m{x}_0$, and secondly the gradients coincide, so $m=1/x_0$. Substituting, $\ln x_0=1$, so $x_0=e$, and $m=1/e$.

If the line $y=mx$ meets the curve at $x=a$ and $x=b$ (with $a<b$), both intersections satisfy $\ln a=ma$ and $\ln b=mb$, so $\frac{\ln a}{a}=\frac{\ln b}{b}=m$. This can be written $\ln (a^{1/a})=\ln (b^{1/b})$, so $a^{1/a}=b^{1/b}$, giving $a^b=b^a$. A sketch of $y=\ln x$ together with lines through the origin shows that for any gradient $m<1/e$, the line meets the curve at two points, one to the left of $e$ and the other to the right: hence $a<e<b$.

Part (ii). If $y=mx+c$ (with $c>0$) meets $y=\ln x$ at $x=p$ and $x=q$ ($p<q$), then $\ln p=mp+c$ and $\ln q=mq+c$, so $\frac{\ln q-\ln p}{q-p}=m$. The gradient $m$ of the chord is strictly less than the gradient of the tangent from the origin ($1/e$), since the line has a positive $y$-intercept ($c>0$): a sketch confirms this, the line being shifted upwards relative to the tangent from the origin. As $m<1/e$, the same reasoning as in (i) (in terms of the function $x\mapsto \ln x/x$) carries over: since $c>0$, the two intersections are closer to $e$ than in the case $c=0$, and the graph confirms that $p<e$ and $q>e$, giving $p^q>q^p$ (comparing $\ln (p^q)=q\ln p$ and $\ln (q^p)=p\ln q$ with the inequality $m=(\ln q-\ln p)/(q-p)$).

Part (iii). The line through $(p,\ln p)$ and $(q,\ln q)$ with $e\le p<q$ has gradient $m=\frac{\ln q-\ln p}{q-p}$. Since $\ln$ is concave and $p\ge e>0$, the gradient of the chord is less than the derivative at $p$, so $m<1/p\le 1/e$. The $y$-intercept is $c=\ln p-mp>0$ (since the concavity of $\ln$ ensures the chord lies below the tangent, but the intercept remains positive for $p\ge e$ — visible on the sketch). The conditions of (ii) are therefore satisfied with this $p$ and $q$, so $p^q>q^p$. Taking $p=e$ and $q=\pi$ (with $e<\pi$): $e^{\pi }>{\pi }^e$.

Part (iv). The condition $\frac{\ln q-\ln p}{q-p}=e^{-1}=\frac{1}{e}$ means that the gradient of the chord is exactly $1/e$, the same as the tangent from the origin in (i). This chord is therefore parallel to the tangent at $x=e$. But the tangent at $x=e$ passes through the origin ($c=0$); a parallel line with non-zero $y$-intercept will have $c<0$ (the line of gradient $1/e$ meeting the curve at two points must be shifted downwards relative to the tangent). With $c<0$, adapting the reasoning of (ii) (reversing the direction of the inequality on $c$) shows that $q^p>p^q$.