How do you find the average rate of change of #f(x) = 4x^3 - 8x^2 - 3# over the interval [-5, 2]?

2 Answers
Jim G.
Aug 29, 2017

#100#

Explanation:

#"the average rate of change is the slope of the secant line"#
#"joining"#

#(-5,f(-5))" and "(2,f(2))#

#"in general "(f(b)-f(a))/(b-a)to[a,b]#

#f(b)=f(2)=32-32-3=-3#

#f(a)=f(-5)=-500-200-3=-703#

#rArr(-3-(-703))/7=100#

Ratnaker Mehta
Aug 29, 2017

# 100.#

Explanation:

The Average Rate of Change of a Function # f : RR to RR,# over

the Interval #[a,b] sub RR# is given by, #{f(b)-f(a)}/(b-a).#

Hence, the reqd. Rate#={f(2)-f(-5)}/{2-(-5)}..............(star).#

# f(x)=4x^3-8x^2-3 rArr f(b)-f(a),#

#=4b^3-8b^2-3-(4a^3-8a^2-3),#

#=4(b^3-a^3)-8(b^2-a^2),#

#=4(b-a)(b^2+ba+a^2)-8(b-a)(b+a),#

#=4(b-a){b^2+ba+a^2-2(b+a)},# so that,

# (f(b)-f(a))/(b-a)=4{b^2+ba+a^2-2(b+a)}, (bnea).#

With, #a=-5, and, b=2,# we have, the desired

#"the desired rate="4{4-10+25-2(-3)}=4*25=100.#

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