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Test solutions College : The Derivative and Its Definition — Zyro the alien explorer of Planète Maths

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Suggested time: 45 minutes. Out of 20 points. Calculator allowed only when the problem says so.

1 Definition of the derivative / 4 pts

(a) \(f(1)=1\); \(f(1+h)=2(1+2h+h^2)-1-h=1+3h+2h^2\). The quotient is \(\dfrac{3h+2h^2}{h}=3+2h\). (2 pts)

(b) As \(h\to 0\), \(3+2h\to 3\), so \(f'(1)=3\). (1 pt)

(c) \(f'(x)=4x-1\), and \(f'(1)=3\). (1 pt)

2 Differentiating / 4 pts

(a) \(g'(x)=20x^3-6x^2+1\). (1 pt)

(b) \(h(x)=3x^{1/2}+4x^{-2}\), so \(h'(x)=\dfrac{3}{2\sqrt{x}}-\dfrac{8}{x^3}\). (2 pts)

(c) Expand: \(k(x)=x^2+4x\), so \(k'(x)=2x+4\). (1 pt)

3 Tangent lines of a cubic / 4 pts

(a) \(f(2)=1\), \(f'(x)=3x^2-4x\), \(f'(2)=4\). Tangent: \(y=1+4(x-2)=4x-7\). (2 pts)

(b) \(3x^2-4x=x(3x-4)=0\) gives \(x=0\) or \(x=\dfrac43\). \(f(0)=1\) and \(f\!\left(\dfrac43\right)=\dfrac{64}{27}-\dfrac{96}{27}+\dfrac{27}{27}=-\dfrac{5}{27}\). Points: \((0,1)\) and \(\left(\dfrac43,-\dfrac{5}{27}\right)\). (2 pts)

4 Filling a tank / 3 pts

(a) \(V(2)=10\), \(V(6)=42\); average rate \(=\dfrac{32}{4}=8\) gal/min. (1 pt)

(b) \(V'(t)=t+4\); \(V'(2)=6\) gal/min and \(V'(6)=10\) gal/min. (1 pt)

(c) The tank fills faster and faster; the average 8 gal/min lies between the instantaneous rates 6 and 10. (1 pt)

5 A differentiable piecewise function / 3 pts

Continuity at 2: \(f(2)=5\), so \(2m+n=5\). (1 pt)

Equal slopes: left derivative \(2x=4\) at 2, so \(m=4\). (1 pt)

Then \(n=5-8=-3\). (1 pt)

6 Continuity and differentiability / 2 pts

(a) True: if \(f'(a)\) exists, \(f(a+h)-f(a)=\dfrac{f(a+h)-f(a)}{h}\cdot h\to f'(a)\cdot 0=0\). (1 pt)

(b) False: \(|x|\) is continuous at \(0\) but its quotients tend to \(1\) and \(-1\). (1 pt)

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