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log(4*x^3+e^x)
  • How to use it?

  • Graphing y =:
  • x^3/2(x+1)^2
  • -x^2-x
  • x/(2x+1)
  • 2x^3-3x^2+5
  • Identical expressions

  • log(four *x^ three +e^x)
  • logarithm of (4 multiply by x cubed plus e to the power of x)
  • logarithm of (four multiply by x to the power of three plus e to the power of x)
  • log(4*x3+ex)
  • log4*x3+ex
  • log(4*x³+e^x)
  • log(4*x to the power of 3+e to the power of x)
  • log(4x^3+e^x)
  • log(4x3+ex)
  • log4x3+ex
  • log4x^3+e^x
  • Similar expressions

  • log(4*x^3-e^x)

Graphing y = log(4*x^3+e^x)

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The graph:

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Intersection points:

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Piecewise:

The solution

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          /   3    x\
f(x) = log\4*x  + e /
$$f{\left(x \right)} = \log{\left(4 x^{3} + e^{x} \right)}$$
f = log(4*x^3 + E^x)
The graph of the function
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:
$$\log{\left(4 x^{3} + e^{x} \right)} = 0$$
Solve this equation
The points of intersection with the axis X:

Numerical solution
$$x_{1} = 0$$
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to log(4*x^3 + E^x).
$$\log{\left(4 \cdot 0^{3} + e^{0} \right)}$$
The result:
$$f{\left(0 \right)} = 0$$
The point:
(0, 0)
Extrema of the function
In order to find the extrema, we need to solve the equation
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
$$\frac{d}{d x} f{\left(x \right)} = $$
the first derivative
$$\frac{12 x^{2} + e^{x}}{4 x^{3} + e^{x}} = 0$$
Solve this equation
Solutions are not found,
function may have no extrema
Inflection points
Let's find the inflection points, we'll need to solve the equation for this
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = $$
the second derivative
$$\frac{24 x - \frac{\left(12 x^{2} + e^{x}\right)^{2}}{4 x^{3} + e^{x}} + e^{x}}{4 x^{3} + e^{x}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = 77.198731330601$$
$$x_{2} = 77.0124850969537$$
$$x_{3} = 60.6525896883197$$
$$x_{4} = 77.0838695174728$$
$$x_{5} = 100$$
$$x_{6} = 64.5261565642568$$
$$x_{7} = 6.04601858923138$$
$$x_{8} = 78$$
$$x_{9} = 70.3757428307212$$
$$x_{10} = 94.25$$
$$x_{11} = 74.3855050604014$$
$$x_{12} = 72.3229314979781$$
$$x_{13} = 51.1071218018637$$
$$x_{14} = 80$$
$$x_{15} = 98$$
$$x_{16} = 52.9946996224697$$
$$x_{17} = 62.5865072888081$$
$$x_{18} = 77.1862034519915$$
$$x_{19} = 96$$
$$x_{20} = 77.8203482568241$$
$$x_{21} = 77.2141744553657$$
$$x_{22} = 56.8055953511733$$
$$x_{23} = 77.3618108426466$$
$$x_{24} = 45.5503908345065$$
$$x_{25} = 54.8948687600961$$
$$x_{26} = 84$$
$$x_{27} = 77.0258369002429$$
$$x_{28} = 82$$
$$x_{29} = 40.2805348535545$$
$$x_{30} = 90$$
$$x_{31} = 47.3809756966331$$
$$x_{32} = 43.7493508424981$$
$$x_{33} = 66.4707529956174$$
$$x_{34} = 77.0566778668501$$
$$x_{35} = 68.419631664462$$
$$x_{36} = 77.1427592390923$$
$$x_{37} = 77.1877987770559$$
$$x_{38} = 86$$
$$x_{39} = 88$$
$$x_{40} = 58.7252674784738$$
$$x_{41} = 77.6251821417523$$
$$x_{42} = 92$$
$$x_{43} = 0$$
$$x_{44} = 41.9873266562552$$
$$x_{45} = 49.234743592684$$

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
$$\left[90, \infty\right)$$
Convex at the intervals
$$\left(-\infty, 0\right]$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty} \log{\left(4 x^{3} + e^{x} \right)} = \infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty} \log{\left(4 x^{3} + e^{x} \right)} = \infty$$
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 log(4*x^3 + E^x), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\log{\left(4 x^{3} + e^{x} \right)}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\log{\left(4 x^{3} + e^{x} \right)}}{x}\right) = 1$$
Let's take the limit
so,
inclined asymptote equation on the right:
$$y = x$$
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:
$$\log{\left(4 x^{3} + e^{x} \right)} = \log{\left(- 4 x^{3} + e^{- x} \right)}$$
- No
$$\log{\left(4 x^{3} + e^{x} \right)} = - \log{\left(- 4 x^{3} + e^{- x} \right)}$$
- No
so, the function
not is
neither even, nor odd
The graph
Graphing y = log(4*x^3+e^x)