MATHEMATICS 180 – Section 005 Midterm Exam #1 – Solutions

1.(20 pts.) Let f(x) = 2x^{3}. Consider the point P = (1, 2) on the graph of f(x).

a) Find the equation of the line L tangent to f(x) at P.

f′(x) = 6x^{2}, f′(1) = 6 = m = slope.
y − 2 = 6(x − 1), or y = 6x − 4.

b) Find points on the graph of f(x) at which the tangent line has slope m = 6. Write down the equations of the tangent lines at these points.

f′(x) = 6x^{2} = 6 = m means x^{2} = 1

which has two solutions x = 1 and x = −1. The first solution gives P_{1} = P = (1, 2) and the tangent was derived in part (a). The second solution gives P_{2} = (−1, −2) and the equation of the tangent at this point reads y + 2 = 6(x + 1).

2. (30 pts.) Compute the following

a) y(x) = (x + 1)^{3} − \frac{1}{x^{2}} , y′(x) = 3(x + 1)^{2} + \frac{2}{x^{3}}

b) \frac{d}{dx}( ^{3} \sqrt{x^{2}} · ^{3} \sqrt{x^{4}}) = \frac{d}{dx}(x^{2}) = 2x.

c) y(x) = (x^{2} + 1)^{4}, y′(x) = 4(x^{2} + 1)^{3}(2x) = 8x(x^{2} + 1)^{3}

d) y(x) = \frac{4}{\sqrt{x^{2}−2x}} = 4(x^{2} − 2x)^{−1/2}, y′(x) = −2(x^{2} − 2x)^{−3/2}(2x − 2).

e) \frac{d}{dx}(2x^{2} − 3) |_{x=5}= (4x) |_{x=5}= 20

f) f(x) = a^{2}x^{2} + 2b^{3}x + c^{4}.

\frac{d}{dx}f(x) = 2a^{2}x + 2b^{3}, \frac{d^{2}}{dx^{2}} f(x) = 2a^{2}.

3. (30 pts.) Compute the following

a) f(P) = (P + 1)^{5}, \frac{d^{2}}{dP^{2}} f(P) = 20(P + 1)^{3}.

b) \frac{d^{2}}{dx^{2}} (x^{3} − x − 1) |_{x=2}= (6x) |_{x=2}= 12.

c) f(x) = (\sqrt{x} + 1)^{3/2}.

f′(x) |_{x=4}= \frac{3}{4}(\sqrt{x} + 1)^{1/2} · x^{−1/2} |_{x=4}= \frac{3\sqrt{3}}{8}.

d) f(x) = x^{10}, \frac{d^{3}}{dx^{3}} f(x) = 720x^{7}.

e) lim_{x→4} \frac{x^{2}−16}{4+x} = 0/8 = 0.

f) lim_{x→2} \sqrt[3]{x^{3} + 19} = ^{3} \sqrt{27} = 3.

4.(20 pts.) Let f(x) = \frac{3}{x}.

a) Compute f(1 + h) − f(1) = \frac{3}{1+h} − 3 = \frac{3}{1+h} − \frac{3(1+h)}{1+h} = \frac{−3h}{1+h}.

b) Compute the difference quotient \frac{f(1+h)−f(1)}{h} = \frac{−3}{1+h}.

c) Compute the limit in b) when h → 0, that is

f′(1) = lim_{h→0} \frac{f(1 + h) − f(1)}{h} = lim_{h→0} \frac{−3}{1 + h} = −3.