Chapter 15 - The Mean-Value Theorem
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The beauty of the Mean-Value Theorem is that you can look at a picture and see that it is true. So here it is:
So now let's state it in words:
| The Mean-Value Theorem |
|---|
| If we make a secant line on $f(x)$ between the points $a$ and $b$, then there is a point in-between $a$ and $b$ (let's call it $e$) where $f\,'(e)$ is the same value as the slope of the secant line. In other words, \[ f\,'(e)=\frac{f(b)-f(a)}{b-a}. \] |
Below you will find a tool to find the secant slope and the point on the function where the derivative equals the secant slope. It defaults to $f(x)=x^3-5$ between $a=-1$ and $b=5$. The blue line is the secant line and the red line is the tangent line.
def _(f=input_box(x^3-5,width=40,label='f(x)='), a=input_box(-1,width=5,label='a='),b=input_box(5,width=5,label='b=')):
aa=min(a,b)
bb=max(a,b)
#------------------
# do the math
df=f.diff(x)
if aa==bb: secant_slope=df(x=aa)
else: secant_slope=(f(x=bb)-f(x=aa))/(bb-aa)
sl=secant_slope*(x-aa)+f(x=aa)
ee=solve(df==secant_slope,x,solution_dict=True)
e=0
for i in range(len(ee)):
# cannot do if aa
if ee[i][x].n()>=aa and bb>=ee[i][x].n():
e=ee[i][x].n()
fe=f(x=e)
dfe=df(x=e)
tl=dfe*(x-e)+fe
#------------------^
#------------------
# print the results
show('f(x)='+latex(f)+r'\mbox{ and }'+'f\'(x)='+latex(df))
en=N(e,digits=4)
fen=N(fe,digits=4)
dfen=N(dfe,digits=4)
secant_slope_n=N(secant_slope,digits=4)
#show('f('+latex(en)+')='+latex(fen)+'\mbox{ and }f\'('+latex(en)+')='+latex(dfen))
show(r'\mbox{Secant Slope=}'+latex(secant_slope_n))
show('e='+latex(en)+r'\mbox{ and }'+'f\'('+latex(en)+')='+latex(dfen))
d=0.05*abs(bb-aa)
domain=[x,aa-d,bb+d]
p=plot(f,domain,thickness=2,color="black")+plot(sl,domain,thickness=2,color="#0099cc")+plot(tl,domain,thickness=2,color="#cc0000")
p.axes_width(1)
p.axes_color((0.3,0.3,0.3))
p.tick_label_color((0.3,0.3,0.3))
p.set_legend_options(loc=(0.15,0.65))
p.legend(false)
p.show(gridlines=True,aspect_ratio='automatic',figsize=5)
#------------------^
Exercise: Use the tool above to find the point where the derivative equals the secant line slope for the following:
- $f(x)=x^3-18x^2+22x-10000$ at $a=-1$ and $b=5$. (Remember to use an * for multiplication and ^ for exponentiation. So you would input x^3-18*x^2+22*x-1000 in the "f(x)=" box.)
- $f(x)=x^2-5$ at $a=-101$ and $b=32$
