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  • Graphing y =:
  • (x-5)/(x-3)
  • x^4+8x^3+16x^2
  • x^4-(1/2)*x^2
  • x²-3x+1
  • Limit of the function:
  • (1+3*x)^(5/x) (1+3*x)^(5/x)

  • Identical expressions

  • (one + three *x)^(five /x)
  • (1 plus 3 multiply by x) to the power of (5 divide by x)
  • (one plus three multiply by x) to the power of (five divide by x)
  • (1+3*x)(5/x)
  • 1+3*x5/x
  • (1+3x)^(5/x)
  • (1+3x)(5/x)
  • 1+3x5/x
  • 1+3x^5/x
  • (1+3*x)^(5 divide by x)

  • Similar expressions

  • (1-3*x)^(5/x)
Plotting a function graph/ (1+3*x)^(5/x)

Graphing y = (1+3*x)^(5/x)

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
                5
                -
                x
f(x) = (1 + 3*x)
f(x)=(3x+1)5xf{\left(x \right)} = \left(3 x + 1\right)^{\frac{5}{x}} 
f = (3*x + 1)^(5/x)
The graph of the function
02468-8-6-4-2-101002000000000000
The domain of the function
The points at which the function is not precisely defined:
x1=0x_{1} = 0
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:
(3x+1)5x=0\left(3 x + 1\right)^{\frac{5}{x}} = 0
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to (1 + 3*x)^(5/x).
(0⋅3+1)50\left(0 \cdot 3 + 1\right)^{\frac{5}{0}}
The result:
f(0)=NaNf{\left(0 \right)} = \text{NaN}
- the solutions of the equation d'not exist
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
(3x+1)5x(15x(3x+1)−5log⁡(3x+1)x2)=0\left(3 x + 1\right)^{\frac{5}{x}} \left(\frac{15}{x \left(3 x + 1\right)} - \frac{5 \log{\left(3 x + 1 \right)}}{x^{2}}\right) = 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
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
5(3x+1)5x(−9(3x+1)2+5(33x+1−log⁡(3x+1)x)2x−6x(3x+1)+2log⁡(3x+1)x2)x=0\frac{5 \left(3 x + 1\right)^{\frac{5}{x}} \left(- \frac{9}{\left(3 x + 1\right)^{2}} + \frac{5 \left(\frac{3}{3 x + 1} - \frac{\log{\left(3 x + 1 \right)}}{x}\right)^{2}}{x} - \frac{6}{x \left(3 x + 1\right)} + \frac{2 \log{\left(3 x + 1 \right)}}{x^{2}}\right)}{x} = 0
Solve this equation
The roots of this equation
x1=43431.2848572218x_{1} = 43431.2848572218
x2=48599.3491396833x_{2} = 48599.3491396833
x3=46531.8841755977x_{3} = 46531.8841755977
x4=47565.5863605614x_{4} = 47565.5863605614
x5=44464.7178283447x_{5} = 44464.7178283447
x6=45498.2563494007x_{6} = 45498.2563494007
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
x1=0x_{1} = 0

lim⁡x→0−(5(3x+1)5x(−9(3x+1)2+5(33x+1−log⁡(3x+1)x)2x−6x(3x+1)+2log⁡(3x+1)x2)x)=2385e154\lim_{x \to 0^-}\left(\frac{5 \left(3 x + 1\right)^{\frac{5}{x}} \left(- \frac{9}{\left(3 x + 1\right)^{2}} + \frac{5 \left(\frac{3}{3 x + 1} - \frac{\log{\left(3 x + 1 \right)}}{x}\right)^{2}}{x} - \frac{6}{x \left(3 x + 1\right)} + \frac{2 \log{\left(3 x + 1 \right)}}{x^{2}}\right)}{x}\right) = \frac{2385 e^{15}}{4}
lim⁡x→0+(5(3x+1)5x(−9(3x+1)2+5(33x+1−log⁡(3x+1)x)2x−6x(3x+1)+2log⁡(3x+1)x2)x)=2385e154\lim_{x \to 0^+}\left(\frac{5 \left(3 x + 1\right)^{\frac{5}{x}} \left(- \frac{9}{\left(3 x + 1\right)^{2}} + \frac{5 \left(\frac{3}{3 x + 1} - \frac{\log{\left(3 x + 1 \right)}}{x}\right)^{2}}{x} - \frac{6}{x \left(3 x + 1\right)} + \frac{2 \log{\left(3 x + 1 \right)}}{x^{2}}\right)}{x}\right) = \frac{2385 e^{15}}{4}
- limits are equal, then skip the corresponding point

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:
Have no bends at the whole real axis
Vertical asymptotes
Have:
x1=0x_{1} = 0
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞(3x+1)5x=1\lim_{x \to -\infty} \left(3 x + 1\right)^{\frac{5}{x}} = 1
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=1y = 1
lim⁡x→∞(3x+1)5x=1\lim_{x \to \infty} \left(3 x + 1\right)^{\frac{5}{x}} = 1
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=1y = 1
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of (1 + 3*x)^(5/x), divided by x at x->+oo and x ->-oo
lim⁡x→−∞((3x+1)5xx)=0\lim_{x \to -\infty}\left(\frac{\left(3 x + 1\right)^{\frac{5}{x}}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞((3x+1)5xx)=0\lim_{x \to \infty}\left(\frac{\left(3 x + 1\right)^{\frac{5}{x}}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
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:
(3x+1)5x=(1−3x)−5x\left(3 x + 1\right)^{\frac{5}{x}} = \left(1 - 3 x\right)^{- \frac{5}{x}}
- No
(3x+1)5x=−(1−3x)−5x\left(3 x + 1\right)^{\frac{5}{x}} = - \left(1 - 3 x\right)^{- \frac{5}{x}}
- No
so, the function
not is
neither even, nor odd
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