You can use pwlf to perform continuous piecewise linear regression in Python. This library can be installed using pip.

There are two approaches in pwlf to perform your fit:

1. You can fit for a specified number of line segments.
2. You can specify the x locations where the continuous piecewise lines should terminate.

Let's go with approach 1 since it's easier, and will recognize the 'gradient change point' that you are interested in.

I notice two distinct regions when looking at the data. Thus it makes sense to find the best possible continuous piecewise line using two line segments. This is approach 1.

import numpy as np
import matplotlib.pyplot as plt
import pwlf

x = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15])
y = np.array([5, 7, 9, 11, 13, 15, 28.92, 42.81, 56.7, 70.59,
              84.47, 98.36, 112.25, 126.14, 140.03])

my_pwlf = pwlf.PiecewiseLinFit(x, y)
breaks = my_pwlf.fit(2)
print(breaks)

[ 1. 5.99819559 15. ]

The first line segment runs from [1., 5.99819559], while the second line segment runs from [5.99819559, 15.]. Thus the gradient change point you asked for would be 5.99819559.

We can plot these results using the predict function.

x_hat = np.linspace(x.min(), x.max(), 100)
y_hat = my_pwlf.predict(x_hat)

plt.figure()
plt.plot(x, y, 'o')
plt.plot(x_hat, y_hat, '-')
plt.show()

The method proposed by Vito M. R. Muggeo[1] is relatively simple and efficient. It works for a specified number of segments, and for a continuous function. The positions of the breakpoints are iteratively estimated by performing, for each iteration, a segmented linear regression allowing jumps at the breakpoints. From the values of the jumps, the next breakpoint positions are deduced, until there are no more discontinuity (jumps).

"the process is iterated until possible convergence, which is not, in general, guaranteed"

In particular, the convergence or the result may depends on the first estimation of the breakpoints.

This is the method used in the R Segmented package.

Here is an implementation in python:

import numpy as np
from numpy.linalg import lstsq

ramp = lambda u: np.maximum( u, 0 )
step = lambda u: ( u > 0 ).astype(float)

def SegmentedLinearReg( X, Y, breakpoints ):
    nIterationMax = 10

    breakpoints = np.sort( np.array(breakpoints) )

    dt = np.min( np.diff(X) )
    ones = np.ones_like(X)

    for i in range( nIterationMax ):
        # Linear regression:  solve A*p = Y
        Rk = [ramp( X - xk ) for xk in breakpoints ]
        Sk = [step( X - xk ) for xk in breakpoints ]
        A = np.array([ ones, X ] + Rk + Sk )
        p =  lstsq(A.transpose(), Y, rcond=None)[0] 

        # Parameters identification:
        a, b = p[0:2]
        ck = p[ 2:2+len(breakpoints) ]
        dk = p[ 2+len(breakpoints): ]

        # Estimation of the next break-points:
        newBreakpoints = breakpoints - dk/ck 

        # Stop condition
        if np.max(np.abs(newBreakpoints - breakpoints)) < dt/5:
            break

        breakpoints = newBreakpoints
    else:
        print( 'maximum iteration reached' )

    # Compute the final segmented fit:
    Xsolution = np.insert( np.append( breakpoints, max(X) ), 0, min(X) )
    ones =  np.ones_like(Xsolution) 
    Rk = [ c*ramp( Xsolution - x0 ) for x0, c in zip(breakpoints, ck) ]

    Ysolution = a*ones + b*Xsolution + np.sum( Rk, axis=0 )

    return Xsolution, Ysolution

Example:

import matplotlib.pyplot as plt

X = np.linspace( 0, 10, 27 )
Y = 0.2*X  - 0.3* ramp(X-2) + 0.3*ramp(X-6) + 0.05*np.random.randn(len(X))
plt.plot( X, Y, 'ok' );

initialBreakpoints = [1, 7]
plt.plot( *SegmentedLinearReg( X, Y, initialBreakpoints ), '-r' );
plt.xlabel('X'); plt.ylabel('Y');

[1]: Muggeo, V. M. (2003). Estimating regression models with unknown breakpoints. Statistics in medicine, 22(19), 3055-3071.