Differential Calculus Questions and Answers

Answer: A

The minimum value is 0.

Find the critical points: \(f'(x) = 6x^{2} - 18x + 12 = 6(x - 1)(x - 2)\), so \(x = 1\) and \(x = 2\). Evaluate at these and at the end points:

\[ \begin{aligned} f(0) &= 0,\quad f(1) = 5 \\ f(2) &= 4,\quad f(3) = 9 \end{aligned} \]

The smallest is \(f(0) = 0\), at the left end point. 4 is only the local minimum at \(x = 2\), and 5 and 9 are the local maximum and the value at \(x = 3\). On a closed interval the end points must always be checked.

Answer: B

The derivative is \(-\dfrac{x}{y^{2}} \ln v - \dfrac{u}{v}\).

By the chain rule, with \(\partial u/\partial y = -x/y^{2}\) and \(\partial v/\partial y = -1\):

\[ \begin{aligned} \frac{\partial F}{\partial y} &= \frac{\partial f}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial y} \\ &= \ln v \left(-\frac{x}{y^{2}}\right) + \frac{u}{v}(-1) \\ &= -\frac{x}{y^{2}}\ln v - \frac{u}{v} \end{aligned} \]

The first option has the right first term but a plus sign on the second, which drops the -1 from \(\partial v/\partial y\). The third and fourth lose the minus sign from differentiating x/y with respect to y.

Answer: D

The correct statement is that \(g(x)\) is differentiable at \(x = 0\).

Near 0, \(\sin x \approx x\), so \(g(x) \approx |x|x\), and

\[ \begin{aligned} g'(0) &= \lim_{h \to 0} \frac{|h|\sin h}{h} \\ &= \lim_{h \to 0} |h|\frac{\sin h}{h} = 0 \end{aligned} \]

from both sides, so the derivative exists and equals 0.

The others fail. \(|f(x)| = |x||\sin x|\), which differs from g wherever \(\sin x \lt 0\). g is odd, the product of the even |x| and the odd sin x. And since \(g = -f\) for \(x \lt 0\), the maxima of g there sit at the minima of f, so the maxima do not coincide.

Answer: A

The true statement is \(f\) is NOT continuous at (0,0).

Partial derivatives first. On the \(x\)-axis, \(f(x,0) = 0\), and on the \(y\)-axis, \(f(0,y) = 0\), so both difference quotients are 0 and \(f_{x}(0,0) = f_{y}(0,0) = 0\). That rules out every option giving a partial derivative the value 1.

Now approach the origin along a path close to the excluded curve \(y = -|x|\), where the denominator vanishes. Take \(x \gt 0\) and \(y = -x + x^{2}\):

\[ \begin{aligned} |x| + y &= x^{2} \\ xy &= -x^{2} + x^{3} \\ f &= -1 + x \to -1 \end{aligned} \]

Along the axes \(f \to 0\), but along this path \(f \to -1\), so the limit at (0,0) does not exist and \(f\) is not continuous there, even though both partial derivatives exist.

Answer: B

The limit is \(\frac{1}{2}\).

Expand both sides of \(f(x + h) - f(x) = hf'(x + \theta h)\) in powers of h, using Taylor's theorem. For the cubic \(f = x^{2} + x^{3}\), \(f''' = 6\) is constant:

\[ \begin{aligned} \text{LHS} &= hf' + \tfrac{h^{2}}{2}f'' + \tfrac{h^{3}}{6}f''' \\ \text{RHS} &= hf' + \theta h^{2}f'' + \tfrac{\theta^{2}h^{3}}{2}f''' \end{aligned} \]

Cancelling \(hf'\) and dividing by \(h^{2}\): \(\tfrac{1}{2}f'' + \tfrac{h}{6}f''' = \theta f'' + \tfrac{\theta^{2}h}{2}f'''\). As \(h \to 0\) this leaves \(\theta f'' = \tfrac{1}{2}f''\), so \(\theta \to \tfrac{1}{2}\) wherever \(f''(x) = 2 + 6x \neq 0\).

The general result is that \(\theta \to 1/2\) for any twice-differentiable f with \(f'' \neq 0\), so 1/3, 1/4 and 4/5 are ruled out.

Answer: D

\((\frac{3}{2}, \frac{3}{4})\) is a saddle point of f.

At \((\frac{3}{2}, \frac{3}{4})\) both first derivatives vanish:

\[ \begin{aligned} f_{x} &= 2xy+2y^{2}-4xy^{2} = 2.25+1.125-3.375 = 0 \\ f_{y} &= x^{2}+4xy-4x^{2}y = 2.25+4.5-6.75 = 0 \end{aligned} \]

The second derivatives there are:

\[ \begin{aligned} f_{xx} &= 2y-4y^{2} = -0.75 \\ f_{yy} &= 4x-4x^{2} = -3 \\ f_{xy} &= 2x+4y-8xy = -3 \end{aligned} \]

\(f_{xx}f_{yy}-f_{xy}^{2} = 2.25 - 9 \lt 0\), so the point is a saddle, not a maximum. At \((\frac{3}{2}, 0)\) and \((0, \frac{3}{4})\) the gradient is not zero (\(f_{y} = 2.25\) and \(f_{x} = 1.125\) respectively), so they are not extreme points at all.