Question #1:
\(y = x^4 - 3x + x^{1/2}\).
\(\displaystyle {dy \over dx} = 4x^3 - 3 + {1 \over 2} x^{-{1/2}}\).
Answer: (c)
Question #2:
\(\displaystyle {\sqrt{1 - {1 \over x}}}\).
\(\displaystyle {dy \over dx}={1 \over {2x^2 {\sqrt{1 - {1 \over x}}}}}\).
Answer: (d)
Question #3:
\(y = 3x^2 - x^3\).
\(y'=6x-3x^2=3x(2-x)\).
\(y(0)=0,\; y(2)=4,\; y(4)=-16\). The maximum is \(y(2)=4\).
Answer: (a)
Question #4:
\(\displaystyle \int_0^1 x^3\,dx
= \left.\frac{x^4}{4}\right|_0^1=\frac{1}{4}-0=\frac{1}{4}\).
Answer: (d)
Question #5:
\(y = x + e^{-x}\).
\(y'=1-e^{-x}>0\) if \(1>e^{-x}\), so \(e^x>1\), okay if \(x>0\).
Answer: (a)
Question #6:
It is an odd function that is positive if \(x>1\) and negative if \(x<-1\).
Answer: (a)
Question #7:
\(x^3 - y^3 + x + y = 10\).
\(\displaystyle 3x^2 - 3y^2 {dy \over dx} + 1 + {dy \over dx} = 0\).
\(\displaystyle 3(2^2) - 3(1)^2 {dy \over dx} + 1
+ {dy \over dx} = 13 - 2{dy \over dx} =0\).
\(\displaystyle {dy \over dx} = {13 \over 2}\).
Answer: (b)
Question #8:
\(A(t) = s(t)^2\) and \(A'(t) = 2s(t)s'(t)\).
\(\displaystyle A'(t_0) = 2 \cdot 10 \cdot {1 \over 4} = 5\) sq. in./sec.
Answer: (c)