Power Functions - Find Where they Equate on Interval

Algebra Level 3

{ f ( x ) = x a + x b + x c + x d g ( x ) = a x + b x + c x + d x \begin{cases} f(x) = x^a + x^b + x^c + x^d \\ g(x) = a^x + b^x + c^x + d^x \end{cases}

Functions f ( x ) f(x) and g ( x ) g(x) are defined as above, where a = 1 3 a=\frac 13 , b = 1 2 b=\frac 12 , c = 1 4 c=\frac 14 , and d = 2 d=2 , are continuous on x [ 0 , 10 ] x \in [0,10] and f ( x ) = g ( x ) f(x)=g(x) at two points x 1 x_1 and x 2 x_2 . Find x 1 + x 2 x_1+x_2 to the nearest 1 / 100 1/100 th.


The answer is 5.54.

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3 solutions

Steven Chase
Apr 25, 2020

Here's a hand-coded Newton Raphson algorithm. The function I call f f is the difference between the two functions given in the problem.

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import math
import random

a = 1.0/3.0
b = 1.0/2.0
c = 1.0/4.0
d = 2.0

#########################################################################################################

x = 10.0*random.random()  # guess an initial value for x

for j in range(0,100):  # Newton-Raphson

    f = x**a + x**b + x**c + x**d - a**x - b**x - c**x - d**x   # function f = 0

    fp = a*(x**(a-1.0)) + b*(x**(b-1.0)) + c*(x**(c-1.0)) + d*(x**(d-1.0))   # derivative
    fp = fp - (a**x)*math.log(a) - (b**x)*math.log(b) - (c**x)*math.log(c) - (d**x)*math.log(d)

    x = x - f/fp  # update guess for x


f = x**a + x**b + x**c + x**d - a**x - b**x - c**x - d**x

print f
print x

#########################################################################################################


#>>> 
#-8.881784197e-16
#0.684878755615
#>>> ================================ RESTART ================================
#>>> 
#-3.5527136788e-15
#4.85210468951
#>>>     

Chew-Seong Cheong
Apr 25, 2020

Using Newton-Raphson method x n + 1 = x n f ( x ) f ( x ) x_{n+1} = x_n - \dfrac {f(x)}{f'(x)} with an Excel spreadsheet. x 1 0.684878756 x_1 \approx 0.684878756 , x 2 4.85210469 x_2 \approx 4.85210469 , and x 1 + x 2 5.536983445 5.54 x_1+x_2 \approx 5.536983445 \approx \boxed{5.54} .

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