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2^x*(4*x-7)
Plotting a function graph/ 2^x*(4*x-7)

Graphing y = 2^x*(4*x-7)

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The graph:

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Intersection points:

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Piecewise:

The solution

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        x          
f(x) = 2 *(4*x - 7)
f(x)=2x(4x−7)f{\left(x \right)} = 2^{x} \left(4 x - 7\right) 
f = 2^x*(4*x - 1*7)
The graph of the function
0-130-120-110-100-90-80-70-60-50-40-30-20-1010-5000050000
The points of intersection with the X-axis coordinate
Graph of the function intersects the axis X at f = 0
so we need to solve the equation:
2x(4x−7)=02^{x} \left(4 x - 7\right) = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=74x_{1} = \frac{7}{4}
Numerical solution
x1=−130.560173892367x_{1} = -130.560173892367
x2=−104.674166405026x_{2} = -104.674166405026
x3=−100.698060959641x_{3} = -100.698060959641
x4=−57.294975833538x_{4} = -57.294975833538
x5=−70.9940544810989x_{5} = -70.9940544810989
x6=−53.4287790344014x_{6} = -53.4287790344014
x7=−55.3579282831068x_{7} = -55.3579282831068
x8=−102.685831208952x_{8} = -102.685831208952
x9=−76.9102986733328x_{9} = -76.9102986733328
x10=−128.567045505463x_{10} = -128.567045505463
x11=−116.61410475038x_{11} = -116.61410475038
x12=1.75x_{12} = 1.75
x13=−118.605484638325x_{13} = -118.605484638325
x14=−124.581558957354x_{14} = -124.581558957354
x15=−110.642193130371x_{15} = -110.642193130371
x16=−78.8861273471272x_{16} = -78.8861273471272
x17=−106.663028005667x_{17} = -106.663028005667
x18=−92.7535477155393x_{18} = -92.7535477155393
x19=−47.7084257679483x_{19} = -47.7084257679483
x20=−98.7108980899914x_{20} = -98.7108980899914
x21=−84.8224250368008x_{21} = -84.8224250368008
x22=−120.597199566848x_{22} = -120.597199566848
x23=−122.589230274903x_{23} = -122.589230274903
x24=−72.9640484285695x_{24} = -72.9640484285695
x25=−80.8635210365604x_{25} = -80.8635210365604
x26=−49.6014171256296x_{26} = -49.6014171256296
x27=−112.632435356791x_{27} = -112.632435356791
x28=−65.0999900791222x_{28} = -65.0999900791222
x29=−108.652380920465x_{29} = -108.652380920465
x30=−126.574169129318x_{30} = -126.574169129318
x31=−96.724389431829x_{31} = -96.724389431829
x32=−63.1418710289193x_{32} = -63.1418710289193
x33=−86.8036898586515x_{33} = -86.8036898586515
x34=−114.62308077281x_{34} = -114.62308077281
x35=−82.8423304777736x_{35} = -82.8423304777736
x36=−90.7693361469761x_{36} = -90.7693361469761
x37=−59.238627125049x_{37} = -59.238627125049
x38=−69.026493050923x_{38} = -69.026493050923
x39=−51.5092035175383x_{39} = -51.5092035175383
x40=−74.9362064115449x_{40} = -74.9362064115449
x41=−61.1878635487672x_{41} = -61.1878635487672
x42=−88.786023529583x_{42} = -88.786023529583
x43=−94.738586806713x_{43} = -94.738586806713
x44=−67.061680055888x_{44} = -67.061680055888
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to 2^x*(4*x - 1*7).
20((−1)7+4⋅0)2^{0} \left(\left(-1\right) 7 + 4 \cdot 0\right)
The result:
f(0)=−7f{\left(0 \right)} = -7
The point:
(0, -7)
Extrema of the function
In order to find the extrema, we need to solve the equation
ddxf(x)=0\frac{d}{d x} f{\left(x \right)} = 0
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
ddxf(x)=\frac{d}{d x} f{\left(x \right)} =
the first derivative
2x(4x−7)log⁡(2)+4⋅2x=02^{x} \left(4 x - 7\right) \log{\left(2 \right)} + 4 \cdot 2^{x} = 0
Solve this equation
The roots of this equation
x1=−4+log⁡(128)4log⁡(2)x_{1} = \frac{-4 + \log{\left(128 \right)}}{4 \log{\left(2 \right)}}
The values of the extrema at the points:
 -4 + log(128)     3/4 /     -4 + log(128)\  -1 
(-------------, 2*2   *|-7 + -------------|*e  )
    4*log(2)           \         log(2)   /


Intervals of increase and decrease of the function:
Let's find intervals where the function increases and decreases, as well as minima and maxima of the function, for this let's look how the function behaves itself in the extremas and at the slightest deviation from:
Minima of the function at points:
x1=−4+log⁡(128)4log⁡(2)x_{1} = \frac{-4 + \log{\left(128 \right)}}{4 \log{\left(2 \right)}}
The function has no maxima
Decreasing at intervals
[−4+log⁡(128)4log⁡(2),∞)\left[\frac{-4 + \log{\left(128 \right)}}{4 \log{\left(2 \right)}}, \infty\right)
Increasing at intervals
(−∞,−4+log⁡(128)4log⁡(2)]\left(-\infty, \frac{-4 + \log{\left(128 \right)}}{4 \log{\left(2 \right)}}\right]
Inflection points
Let's find the inflection points, we'll need to solve the equation for this
d2dx2f(x)=0\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
d2dx2f(x)=\frac{d^{2}}{d x^{2}} f{\left(x \right)} =
the second derivative
2x((4x−7)log⁡(2)+8)log⁡(2)=02^{x} \left(\left(4 x - 7\right) \log{\left(2 \right)} + 8\right) \log{\left(2 \right)} = 0
Solve this equation
The roots of this equation
x1=−8+log⁡(128)4log⁡(2)x_{1} = \frac{-8 + \log{\left(128 \right)}}{4 \log{\left(2 \right)}}

Сonvexity and concavity intervals:
Let’s find the intervals where the function is convex or concave, for this look at the behaviour of the function at the inflection points:
Concave at the intervals
[−8+log⁡(128)4log⁡(2),∞)\left[\frac{-8 + \log{\left(128 \right)}}{4 \log{\left(2 \right)}}, \infty\right)
Convex at the intervals
(−∞,−8+log⁡(128)4log⁡(2)]\left(-\infty, \frac{-8 + \log{\left(128 \right)}}{4 \log{\left(2 \right)}}\right]
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞(2x(4x−7))=0\lim_{x \to -\infty}\left(2^{x} \left(4 x - 7\right)\right) = 0
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=0y = 0
lim⁡x→∞(2x(4x−7))=∞\lim_{x \to \infty}\left(2^{x} \left(4 x - 7\right)\right) = \infty
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of 2^x*(4*x - 1*7), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(2x(4x−7)x)=0\lim_{x \to -\infty}\left(\frac{2^{x} \left(4 x - 7\right)}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(2x(4x−7)x)=∞\lim_{x \to \infty}\left(\frac{2^{x} \left(4 x - 7\right)}{x}\right) = \infty
Let's take the limit
so,
inclined asymptote on the right doesn’t exist
Even and odd functions
Let's check, whether the function even or odd by using relations f = f(-x) и f = -f(-x).
So, check:
2x(4x−7)=2−x(−4x−7)2^{x} \left(4 x - 7\right) = 2^{- x} \left(- 4 x - 7\right)
- No
2x(4x−7)=−2−x(−4x−7)2^{x} \left(4 x - 7\right) = - 2^{- x} \left(- 4 x - 7\right)
- No
so, the function
not is
neither even, nor odd
The graph
Graphing y = 2^x*(4*x-7)
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