Practice Problems

Work through the odd-numbered problems 1-17. Once you have completed the problem set, check your answers.

1. 

 x
 		 y = f(x)
 m(x)  =  the estimated slope of the tangent line to y=f(x) at the point (x,y)
 0  1 1
         0.5 1.4 1/2
         1.0 1.6 0
         1.5 1.4 –1/2
         2.0 1 –2
         2.5 0 –2
         3.0 –1 –2
         3.5 –1.3 0
         4.0 –1 1

3. 

(a) At x=1,3, and 4 .

(b) \mathrm{f} is largest at \mathrm{x}=4 . \mathrm{f} is smallest at \mathrm{x}=3

5. (a) Graph          (b) Graph          (c) \mathrm{m}(\mathrm{x})=\cos (\mathrm{x})

7. The solution is similar to the method used in Example 4. Assume we turn off the engine at the point (p, q) on the curve y=x^{2}, and then find values of p and q so the tangent line to y=x^{2} at the point (p, q) goes through the given point (5,16) .(p, q) is on y=x^{2} so q=p^{2}. The equation of the tangent line to y=x^{2} at \left(p, p^{2}\right) is y=2 p x-p^{2} so, substituting x=5 and y=16, we have 16=2 p(5)-p^{2} Solving \mathrm{p}^{2}-10 \mathrm{p}+16=0 we get \mathrm{p}=2 or \mathrm{p}=8. The solution we want (moving left to right along the curve) is \mathrm{p}=2, \mathrm{q}=\mathrm{p}^{2}=4 (\mathrm{p}=8, \mathrm{q}=64 would be the solution if we were moving right to left).

9. Impossible. The point (1,3) is "inside" the parabola.

11.

(a) \mathrm{m}_{\mathrm{sec}}=\frac{\mathrm{f}(\mathrm{x}+\mathrm{h})-\mathrm{f}(\mathrm{x})}{(\mathrm{x}+\mathrm{h})-(\mathrm{x})}=\frac{\{3(\mathrm{x}+\mathrm{h})-7\}-\{3 \mathrm{x}-7\}}{(\mathrm{x}+\mathrm{h})-\mathrm{x}}=\frac{3 \mathrm{~h}}{\mathrm{~h}}=3

(b) \mathrm{m}_{\tan }=\lim\limits_{h \rightarrow 0} m_{\mathrm{sec}}=\lim\limits_{h \rightarrow 0} 3=3.

(c) At \mathrm{x}=2, \mathrm{~m}_{\tan }=3

(d) f(2)=-1 so the tangent line is y-(-1)=3(x-2) or y=3 x-7

13.

(a) m_{\sec }=\frac{f(x+h)-f(x)}{(x+h)-(x)}=\frac{\{a(x+h)+b\}-\{a x+b\}}{(x+h)-x}=\frac{a h}{h}=a

(b) \mathrm{m}_{\mathrm{tan}}=\lim\limits_{h \rightarrow 0} m_{\mathrm{sec}}=\lim\limits_{h \rightarrow 0} a=a

(c) At \mathrm{x}=2, \mathrm{~m}_{\mathrm{tan}}=\mathrm{a}.

(d) f(2)=2 a+b so the tangent line is y-(2 a+b)=a(x-2) or y=a x+b.

15.

(a) m_{\mathrm{sec}}=\frac{\mathrm{f}(\mathrm{x}+\mathrm{h})-\mathrm{f}(\mathrm{x})}{(\mathrm{x}+\mathrm{h})-(\mathrm{x})}=\frac{\left\{8-3(\mathrm{x}+\mathrm{h})^{2}\right\}-\left\{8-3 \mathrm{x}^{2}\right\}}{(\mathrm{x}+\mathrm{h})-\mathrm{x}}=\frac{-6 \mathrm{xh}-3 \mathrm{~h}^{2}}{\mathrm{~h}}=-6 \mathrm{x}-3 \mathrm{~h}

(b) \mathrm{m}_{\mathrm{tan}}=\lim\limits_{h \rightarrow 0} m_{\mathrm{sec}}=\lim\limits_{h \rightarrow 0}-6 x-3 h=-6 x \quad 

(c) At \mathrm{x}=2, \mathrm{~m}_{\mathrm{tan}}=-6(2)=-12.

(d) \mathrm{f}(2)=-4 so the tangent line is \mathrm{y}-(-4)=-12(\mathrm{x}-2) or \mathrm{y}=-12 \mathrm{x}+20 \ldots

17. \mathrm{a}=1, \mathrm{~b}=2, \mathrm{c}=0, so \mathrm{m}_{\mathrm{tan}}=(2)(1)(\mathrm{x})+2=2 \mathrm{x}+2. The problem is to find \mathrm{p} for which 6-\left(p^{2}+2 p\right)=(2 p+2)(3-p)

This reduces to p^{2}-6 p=0 so p=0 or 6 and the required points are (0,0) and (6,48).