Cumulative assessment: Calculus I

From input replacement to a point and a slope

For any real $a$ and $h\ne0$,

Taking $h\to0$ gives $\boxed{f'(a)=4-2a}$. At input $1$, $f(1)=3$ and $f'(1)=2$, so the tangent line is $y-3=2(x-1)$, or $\boxed{y=2x+1}$. The quotient requires $h\ne0$; the derivative’s limiting value does not.

A standard trigonometric limit controls the quotient

Using the exact identity $1-\cos3x=2\sin^2(3x/2)$ gives

The underlying sine limit uses radians. Alternatively the original quotient is $0/0$, with differentiable numerator and denominator and nonzero denominator derivative for $x\ne0$. L’H^opital gives $3\sin3x/(2x)$, again a $0/0$ quotient with denominator derivative $2\ne0$; a second application gives $9\cos3x/2\to9/2$. In either method, the value is justified by a limit theorem, not by setting $0/0$ equal to a number.

First match heights, then compare limiting slopes

The left limit at $1$ is $1$ and the right limit is also $1$. Therefore continuity requires and is obtained by $\boxed{k=1}$. With that value, the left difference quotient is $[(1+h)^2-1]/h=2+h$ for $h<0$; the right quotient is $[2(1+h)-1-1]/h=2$ for $h>0$. Both tend to $2$, so $\boxed{g'(1)=2}$. A piecewise formula can be differentiable at its joining point when both height and slope agree.

Determine positivity before expanding logarithms

The numerator $x^2+1$ is strictly positive. Its quotient by $x+2$ is positive exactly when $x>-2$, so the real domain is $\boxed{(-2,\infty)}$. On that domain both logarithms are defined and $f=\ln(x^2+1)-\ln(x+2)$. Hence

The derivative formula may have values elsewhere, but they are not derivatives of this real logarithm expression there because the original function is not defined.

Differentiate the equation, including both parts of a product

Implicit differentiation gives $2x+2(xy'+y)+6yy'=0$. Collecting the $y'$ terms produces $(2x+6y)y'=-(2x+2y)$. The point $(1,2)$ satisfies the original equation, and the coefficient $2x+6y$ equals $14\ne0$ there, so division is legitimate. The slope is $\boxed{-3/7}$, and the tangent line is $\boxed{y-2=-\tfrac37(x-1)}$. One can also confirm the local branch without multivariable notation: the quadratic formula gives $y=(-x+\sqrt{51-2x^2})/3$ near $x=1$. Its radicand is $49$ at that point, so this branch is differentiable nearby and passes through $(1,2)$. Differentiating it at $1$ checks the same slope.

Differentiate the geometric relation with respect to time

For a sphere, $V=4\pi r^3/3$. Since $r$ varies with time, $\dd V/\dd t=4\pi r^2(\dd r/\dd t)$. Substitute the stated instant only after differentiating: $12\pi=4\pi(2)^2r'$, so $\boxed{r'=3/4\text{ cm/s}}$. Dividing volume rate by the sphere’s surface area gives length per time, checking the units.

Build an objective on the feasible domain

Let $x>0$ be the horizontal coordinate of the upper-right corner. Width is $2x$, height is $12-x^2$, and feasibility requires $0<x<\sqrt{12}$. The area is $A=2x(12-x^2)=24x-2x^3$, with derivative $A'=24-6x^2$. Its only feasible zero is $x=2$. The derivative is positive before $2$ and negative after; also area tends to zero at either feasible endpoint. Therefore the global maximum has $\boxed{\text{width }4,\ \text{height }8,\ \text{area }32}$ in consistent units. Confusing $x$ with the full width would omit a factor of two from the geometry.

The theorem’s hypotheses are part of its use

The function is undefined at zero, so it is not continuous on the whole closed interval and not differentiable on its whole interior. The Mean Value Theorem cannot be used. The endpoint secant slope is indeed $(1-(-1))/(1-(-1))=1$, but $f'(x)=-1/x^2<0$ at every point of its domain, so no derivative can equal $1$. The missing point allows the endpoint comparison to cross disconnected branches of the graph.

Distance requires a sign split

Factor $v=(t-2)(t-4)$: velocity is positive on $(0,2)$, negative on $(2,4)$, and positive on $(4,5)$. Use $F(t)=t^3/3-3t^2+8t$, so $F(0)=0$, $F(2)=20/3$, $F(4)=16/3$, and $F(5)=20/3$. Displacement is $F(5)-F(0)=\boxed{20/3}$ m. Distance adds magnitudes of the three changes:

Average velocity is displacement divided by five seconds, $\boxed{4/3}$ m/s. The particle has traveled farther than its net position change because it reverses direction twice.

FTC supplies a derivative without an elementary antiderivative

The integrand is continuous for every real $t$, so FTC and the chain rule give the exact identity $\boxed{A'(x)=2x\sqrt{1+x^8}}$. At $x=1$, the integration interval has zero length, so $A(1)=0$, and the derivative is $2\sqrt2$. The linearization is $L(x)=2\sqrt2(x-1)$, giving $\boxed{A(1.01)\approx0.02\sqrt2}$. The relation $A'(x)=2x\sqrt{1+x^8}$ is exact; replacing $A(1.01)$ with $L(1.01)$ is an approximation whose error has not been asserted to be zero.

The same region generates two different integrands

Intersections satisfy $4-x^2=x+2$, or $(x+2)(x-1)=0$, so the limits are $-2$ and $1$. The parabola is above the line between them. Area is

Both functions are nonnegative on this interval, so rotation about the $x$-axis gives washers with outer radius $4-x^2$ and inner radius $x+2$. Thus

Area uses a difference of heights; washer volume uses a difference of squared radii, not the square of their difference.

A Newton iterate and an independently justified error bound

Newton’s update gives $x_1=2-(2^3-5)/(3\cdot2^2)=\boxed{7/4}$. Continuity and $f(1)=-4<0<f(2)=3$ give a root in $(1,2)$ by IVT. Because $f'=3x^2>0$ throughout $[1,2]$, the function is strictly increasing there, so the root is unique. Let $a=7/4$. The residual is $f(a)=343/64-5=23/64$. By MVT between $a$ and $r$, $|f(a)-f(r)|=|f'(c)|\,|a-r|$ for some $c\in(1,2)$, where $|f'(c)|\ge3$. Therefore

This bound is conservative but rigorous. The small residual alone would not quantify root error without control of the derivative. Newton’s method does not converge from every starting point; this separate estimate assesses this particular iterate.