Calculus mixed test 1

1. 1
   Factor

   Use the difference-of-cubes identity.

   $$x^3-8=(x-2)(x^2+2x+4)$$
2. 2
   Cancel and evaluate

   The simplified polynomial is continuous at 2.

   $$2^2+2(2)+4=12$$

Independent check: The limit is the derivative of x³ at x = 2, which is 3(2²) = 12.

1. 1
   Left-hand limit

   Use 2x below 1.

   $$\lim_{x\to1^-}f(x)=2$$
2. 2
   Right-hand limit

   Use x plus 2 at and above 1.

   $$\lim_{x\to1^+}f(x)=3$$
3. 3
   Compare

   A two-sided limit needs equality.

   $$2\ne3\Rightarrow\mathrm{DNE}$$

Independent check: Values sampled immediately to the left remain near 2 while right-side values remain near 3.

1. 1
   Apply the power rule

   Multiply each coefficient by its exponent.

   $$(2x^4)'=8x^3,\quad(-5x^2)'=-10x$$
2. 2
   Combine

   The derivative of 3x is 3 and the derivative of -1 is zero.

   $$f'(x)=8x^3-10x+3$$

Independent check: A finite difference at x = 1 approaches 1, matching the formula value 8 - 10 + 3.

1. 1
   Apply the product rule

   Pair each derivative with the untouched other factor.

   $$g'(x)=2x\ln x+(x^2+1)\frac1x$$
2. 2
   State the domain

   The original logarithm requires positive x.

   $$g'(x)=2x\ln x+\frac{x^2+1}{x},\quad x>0$$

Independent check: Expanding the second term to x + 1/x yields an equivalent derivative on x > 0.

1. 1
   Rewrite as a power

   Use exponent one half.

   $$h(x)=(1+5x)^{1/2}$$
2. 2
   Differentiate outside and inside

   Multiply by the derivative of 1 + 5x.

   $$h'(x)=\frac12(1+5x)^{-1/2}(5)$$

Independent check: Squaring h(x) gives 1 + 5x; implicit differentiation of h² gives 2hh′ = 5.

1. 1
   Find an antiderivative

   Integrate the linear function term by term.

   $$F(x)=x^2+x$$
2. 2
   Evaluate

   Subtract the lower endpoint value.

   $$F(3)-F(1)=12-2=10$$

Independent check: The trapezoid under y = 2x + 1 has width 2 and average height 5, so its area is 10.

1. 1
   Substitute

   Use the inner cubic.

   $$u=x^3,\quad du=3x^2dx$$
2. 2
   Integrate

   The transformed integral is twice cosine u.

   $$2\int\cos u\,du=2\sin u+C$$
3. 3
   Return

   Replace u with x cubed.

   $$2\sin(x^3)+C$$

Independent check: The chain rule differentiates the result to 6x² cos(x³).

1. 1
   Evaluate the integrand

   Replace t with x squared.

   $$(x^2)^3+1=x^6+1$$
2. 2
   Differentiate the bound

   Multiply by the derivative of x squared.

   $$Q'(x)=2x(x^6+1)$$

Independent check: Evaluating Q explicitly and differentiating yields the same polynomial.

1. 1
   Find the point

   Evaluate the original function.

   $$f(2)=4-6+1=-1$$
2. 2
   Find the slope

   Differentiate and evaluate.

   $$f'(x)=2x-3,\quad f'(2)=1$$
3. 3
   Write the line

   Use point-slope form.

   $$y+1=x-2\Rightarrow y=x-3$$

Independent check: The line passes through (2, -1), and its slope matches f′(2) = 1.

1. 1
   Reduce to one variable

   Use the perimeter constraint.

   $$W=12-L,\quad A(L)=12L-L^2$$
2. 2
   Find the critical point

   Set the derivative equal to zero.

   $$A'(L)=12-2L=0\Rightarrow L=6$$
3. 3
   Confirm the maximum

   The second derivative is negative and W is also 6.

   $$A''(L)=-2,\quad A=36$$

Independent check: Completing the square gives A = 36 - (L - 6)², which cannot exceed 36.