If \(f(g(x)) = 5\) and \(f(x) = x + 3\) for all real \(x\), then find \(g(x)\).
This question is about composition of functions where \(f(g(x)) = (f \circ g)(x)\).
This means that something goes into \(g(x)\) and then the result of \(g(x)\) goes into \(f(x)\), so suppose that \(y = g(x)\), then \(f(y) = 5 = y + 3\). We can solve for \(y\) yielding \(g(x) = 2\) for all \(x\). (d)
Evaluate
\[ \lim_{x \rightarrow 0} \frac{\tan x}{\cos x}\]From properties of limits we should remember that the limit of a quotient is the quotient of the limits. I literally only remember three points of the tangent function. \(\tan{0} = 0\), \(\tan{\pi/4} = 1\) and \(\tan{\pi/2}\) is undefined. Since \(\tan{0} = 0\) and \(\cos{0} = 1\), the limit here is \(0\). (c)
This is a pretty simple integral to compute since the integral of the exponental function is just itself. In fact, it looks like the concepts being checked here are basic integrals and properties of logarithms and exponents.
\[\begin{aligned}\int_{0}^{\log{4}} \exp{2x}\ dx &= \frac{1}{2}\exp{2x} \Big\vert_{0}^{\log{4}}\\ &= \frac{1}{2}\Big( \exp{(2\log{4}) - \exp{(0)}} \Big) \end{aligned}\]By properties of logs, \(a \log b = \log(b^a)\), and \(\exp 0 = 1\). Therefore the result here is,
\[ \frac{1}{2}\exp(\log 4^2) - \frac{1}{2} = 8 - \frac{1}{2} = \frac{15}{2}\,. \]The answer is therefore (a).
Let \(A - B\) denote \(\{ x \in A : x \notin B \}\). If \((A - B) \cup B = A\), which is true:
(a) \(B\) is empty,
(b) \(A \subseteq B\),
(c) \(B \subseteq A\),
(d) \((B - A) \cup A = B\), or
(e) none of the above.
This emphasizes the importance of reading all of the options before picking one, because I almost got this one wrong.
Technically, (a) is true because if \(B\) is empty, then the equation is true. Let \(A = \{ a_1, a_2, \dots a_n \}\) and \(B = \{ \}\). \(A - B = \{ a_1, a_2, \dots a_n \}\) and the union of a non-empty set with an empty set is just going to yield the non-empty set, so \((A - \{ \}) \cup \{\} = A\). However, (c) is more true, because it is also true if \(B \subseteq A\) because \(\{ \} \subseteq A\) if \(A\) is non-empty and it's also true if \(B\) is non-empty and is indeed a subset of \(A\). Therefore the answer is (c).
If \(f(x) = |x| + 3x^2\) for all real \(x\), then what is \(f^\prime(-1)\)?
This is pretty easy too, we just need to take the derivative of \(f(x)\) and plug in \(-1\). It's a simple enough derivative too, so there's no need to utilize any shortcuts.
\[ f^\prime(x) = \frac{x}{|x|} + 6x \,\]therefore, \(f^\prime(-1) = -1 - 6 = -7\) which is (a).
For what value of \(b\) is the value of \(\int_b^{b+1} (x^2 + x)\ dx\) a minimum?
Generally, how do we find where a function is a minimum (or maximum)? We identify where the derivative is \(0\), right?
Let \(f(x) = x^2 + x\) and \(F(x)\) is the integral of \(f(x)\). Then our result becomes \(F(b+1) - F(b)\). Since \(f(x) = F^\prime(x)\), then we just need to look at where \(F^\prime(b+1) - F^\prime(b) = 0\). Since differentiation and integration are linear operators, a little bit of algebra is all we need to do to find \(b\).
\[\begin{aligned}0 &= f(b+1) - f(b) \\ &= (b+1)^2 + (b+1) - b^2 - b \\ &= b^2 + 2b + 1 + b + 1 - b^2 - b \\ &= 2b + 2 \\ &= 2(b + 1)\\ \\ \therefore\ \ b &= -1 \end{aligned}\]In how many of the eight standard octants of \(xyz\)-space does the graph of \(z = e^{x + y}\) appear?
This problem requires thinking about the bounds of this function and we don't even need to take any official limits. As \(x\) and \(y\) become more and more negative, they both approach \(0\) and as they become more and more positive, they both approach \(+\infty\). Just thinking about how the function will spread out, it should be pretty obvious that it will never expand below \(z = 0\).
Therefore, the answer is \(4\) or (d).
Suppose that the function \(f\) is defined on an interval by the formula \(f(x) = \sqrt{\tan^2(x) - 1}\). If \(f\) is continuous, which of the following intervals could be its domain?
We haven't been told that we are working in \(\mathbb{R}\), but it's a good assumption given the options.
I only know a couple of points of \(\tan\) off the top of my head, but I do know that \(\tan \pi/4 = 1\) which means that \(\tan -\pi/4 = -1\). \(-1\) is fine since we're squaring the result, we just can't have anything smaller than \(|1|\). For \(x < \pi/4\), \(\tan x\) will get smaller, which won't work, so \(\pm\pi/4\) needs to be a maximum which means we must need to cycle counter-clockwise on the unit circle. I solved this problem by just thinking through what could be a valid interval and it turns out that the interval we want is \((\pi/4 , \pi/2 )\) or (b).
I imagine that this problem is evaluating basic integration methods, specifically \(u\)-substitution. Just looking at this integral, when we take the derivative of \(2 - x^2\), we'll have our left over \(x\) and the integration should be pretty painless.
Let \(u = 2 - x^2\) and \(du = - 2x\ dx\) which means that we need to add a factor of \(-\frac{1}{2}\). When \(x = 0\) then \(u = 2\) and when \(x = 1\), then \(u = 1\).
\[\begin{aligned} \int_0^1 \frac{x}{2 - x^2}\ dx &= -\frac{1}{2}\int \frac{1}{u}\ du \\ &= -\frac{1}{2}\log{u}\Big\vert_2^1 \\ &= -\frac{1}{2}\Big( \log(1) - \log(2) \Big) \\ &= \boxed{\frac{\log 2}{2}} \end{aligned}\]Or (e)
If \(f^{\prime\prime}(x) = f^\prime(x)\) for all real \(x\) and if \(f(0) = 0\) and \(f^\prime(0) = -1\), then \(f(x)\) is what?
This is one of those GRE questions which are solved by process of elimination.
The options are:
(a) \(1-e^x\)
(b) \(e^x - 1\)
(c) \(e^{-x} - 1\)
(d) \(e^{-x}\)
(e) \(-e^x\)
Since we know that \(e^0 = 1\) and \(f(0) = 0\), that immediately means (d) and (e) are not possible. Since \(f^\prime(x) = -1\), then (b) is not possible. Therefore, it must be (a) or (c). That being said, it must be (a) because (c) will keep flipping signs with each derivative so the second derivative can't be equal to the first. Therefore it must be (a).
If \(\phi(x, y, z) = x^2 + 2xy + xz^{3/2}\), which of the derivatives are equal to \(0\)?
Our options are:
\(\frac{\partial^2\phi}{\partial y^2}\)
\(\frac{\partial^2\phi}{\partial x \partial y}\)
\(\frac{\partial^2\phi}{\partial z \partial y}\)
At quick glance, looks like the first and last to me since the second will leave some things left over, but first and third will clear it all out based on the combinations of terms.