Prerequisite diagnostic and repair map

Use common-sized fractional pieces

The least common denominator of $10$ and $4$ is $20$. Thus $7/10=14/20$ and $1/4=5/20$, so the difference is $\boxed{9/20}$. Subtracting denominators would change the size of the pieces while subtracting their counts; that is not how fractions represent quantities. A decimal check gives $0.70-0.25=0.45$, consistent with $9/20$.

Expand a square and subtract an entire expression

First $(a-3)^2=a^2-6a+9$. Then

The subtraction changes $-4a$ to $+4a$ and $+5$ to $-5$. Substituting $a=0$ gives $13$ on both sides as a quick arithmetic check; the distribution itself establishes the identity for all real $a$.

Cancel a factor, not a term

Factor the difference of squares: $x^2-16=(x-4)(x+4)$. The original denominator requires $x\ne4$. On that domain, $\frac{(x-4)(x+4)}{x-4}=\boxed{x+4\text{ for }x\ne4}$. The simplified formula would have a value at $4$, but the original fraction does not. Cancellation does not fill the original hole.

Record excluded inputs before clearing denominators

Require $x\ne-1,2$. Multiply by $(x+1)(x-2)$ to obtain $3(x-2)=2(x+1)$. Distribute: $3x-6=2x+2$, hence $\boxed{x=8}$. It is allowed, and the original sides are $3/9=1/3$ and $2/6=1/3$.

A product is negative when its signs differ

The factors vanish at $-2$ and $1$. For $x<-2$, both are negative and their product is positive. Between $-2$ and $1$, $x-1<0$ and $x+2>0$, so the product is negative. For $x>1$, both are positive. The inequality is strict, so omit the zeros: $\boxed{(-2,1)}$.

Replace every input by the whole new expression

Substituting $-3$ gives $f(-3)=(-3)^2+2(-3)-5=9-6-5=\boxed{-2}$. Substituting $x+h$ instead gives

It is not generally $f(x)+h$. For example $x=0,h=2$ gives $f(2)=3$ but $f(0)+2=-3$. One counterexample disproves the proposed identity.

Intersect all input restrictions

The square root requires $5-x\ge0$, or $x\le5$. The denominator requires $x\ne-2$. Both must hold, giving $\boxed{(-\infty,-2)\cup(-2,5]}$. The endpoint $5$ is allowed because a square root of zero is defined and the denominator there is nonzero.

An input condition and an output bound

The radicand requires $x+1\ge0$, so the domain is $\boxed{[-1,\infty)}$. Since $\sqrt{x+1}\ge0$, the output is at most $3$. It equals $3$ at $x=-1$ and decreases without bound as $x$ grows, so the range is $\boxed{(-\infty,3]}$. A domain lists allowed inputs; a range lists achieved outputs.

Composition order changes the operation order

For $f\circ g$, subtract $3$ first and then square: $\boxed{(x-3)^2}$. For $g\circ f$, square first and then subtract $3$: $\boxed{x^2-3}$. They are not the same function; at zero their values are $9$ and $-3$.

Use a logarithm identity only on positive arguments

The domain requires $x-1>0$. Since $4,2>0$, the right side is $\ln8$. The natural logarithm is one-to-one on positive inputs, so $x-1=8$ and $\boxed{x=9}$. This satisfies $x>1$ and returns $\ln8$ in the original equation.

Reference angle and quadrant determine the signs

$5\pi/6=\pi-\pi/6$ lies in quadrant II. Its reference angle is $\pi/6$, whose sine is $1/2$ and cosine magnitude is $\sqrt3/2$. In quadrant II, sine is positive and cosine is negative: $\boxed{\sin(5\pi/6)=1/2,\quad\cos(5\pi/6)=-\sqrt3/2}$. The values satisfy $\sin^2\theta+\cos^2\theta=1$.

Find a slope, then attach it to a point

The slope is $m=(-2-4)/(3-1)=-6/2=-3$. Point-slope form using $(1,4)$ is $y-4=-3(x-1)$, so $\boxed{y=-3x+7}$. At $x=1$ it gives $4$; at $x=3$ it gives $-2$. The slope $-3$ describes output change per unit input, while the intercept $7$ is the value at input zero.

The endpoints determine average velocity, not the whole path

On $[1,3]$, average velocity is $(13-5)/(3-1)=\boxed4$ m/s. On $[1,5]$, it is $(9-5)/(5-1)=\boxed1$ m/s. The table does not rule out additional reversals between measurements, so it does not determine total distance. It forces at least $|13-5|+|9-13|=12$ m of travel, but the actual distance could be larger.

Keep the original nonzero interval until after cancellation

First write

Expanding and distributing the subtraction gives

Since $h\ne0$, the common factor can be canceled. The result is $\boxed{2x+h+2}$, still a finite-interval average rate. Taking $h\to0$ is a subsequent step, not permission to put zero in the original denominator.

A point value does not control the nearby inputs

The definition explicitly gives $\boxed{f(1)=9}$. For every nearby input other than $1$, the value is $x+2$, which approaches $3$ as $x\to1$. Therefore $\boxed{\lim_{x\to1}f(x)=3}$. The limit’s definition examines a punctured neighborhood, excluding the input $1$ itself. Changing only that single point cannot change this nearby limit.

Separate the height from the slope

$f(3)=2$ locates the point $(3,2)$. The derivative value $f'(3)=-4$ is the tangent slope at that input. Thus

Substituting $x=3$ returns $2$, and the coefficient of $x$ is the required slope. The intercept is $14$, not the given point height $2$.