Lösung 1.3:3e
Aus Online Mathematik Brückenkurs 2
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<li>We obtain the critical points by setting the derivative equal to zero, | <li>We obtain the critical points by setting the derivative equal to zero, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
f^{\,\prime}(x) &= (x^2-x-1)'e^x + (x^2-x-1)\bigl(e^x\bigr)^{\prime}\\[5pt] | f^{\,\prime}(x) &= (x^2-x-1)'e^x + (x^2-x-1)\bigl(e^x\bigr)^{\prime}\\[5pt] | ||
&= (2x-1)e^x + (x^2-x-1)e^x\\[5pt] | &= (2x-1)e^x + (x^2-x-1)e^x\\[5pt] | ||
| Zeile 18: | Zeile 18: | ||
This expression for the derivative can only be zero when <math>x^2+x-2=0</math>, because <math>e^x</math> differs from zero for all values of <math>x</math>. We solve the second-degree equation by completing the square, | This expression for the derivative can only be zero when <math>x^2+x-2=0</math>, because <math>e^x</math> differs from zero for all values of <math>x</math>. We solve the second-degree equation by completing the square, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\Bigl(x+\frac{1}{2}\Bigr)^2 - \Bigl(\frac{1}{2}\Bigr)^2 - 2 &= 0\,,\\[5pt] | \Bigl(x+\frac{1}{2}\Bigr)^2 - \Bigl(\frac{1}{2}\Bigr)^2 - 2 &= 0\,,\\[5pt] | ||
\Bigl(x+\frac{1}{2}\Bigr)^2 &= \frac{9}{4}\,,\\[5pt] | \Bigl(x+\frac{1}{2}\Bigr)^2 &= \frac{9}{4}\,,\\[5pt] | ||
| Zeile 37: | Zeile 37: | ||
We can factorize the derivative somewhat, | We can factorize the derivative somewhat, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>f^{\,\prime}(x) = (x^2+x-2)e^x = (x+2)(x-1)e^x\,,</math>}} |
since <math>x^2+x-2</math> has zeros at <math>x=-2</math> and <math>x=1</math>. Each individual factor in the derivative has a sign that is given in the table: | since <math>x^2+x-2</math> has zeros at <math>x=-2</math> and <math>x=1</math>. Each individual factor in the derivative has a sign that is given in the table: | ||
Version vom 12:56, 10. Mär. 2009
As always, a function can only have local extreme points at one of the following types of points,
- critical points, i.e. where
f ,
(x)=0 - points where the function is not differentiable, and
- endpoints of the interval of definition.
We investigate these three cases.
- We obtain the critical points by setting the derivative equal to zero,
f (x)=(x2−x−1)ex+(x2−x−1)
ex
=(2x−1)ex+(x2−x−1)ex=(x2+x−2)ex.This expression for the derivative can only be zero when
x2+x−2=0 , becauseex differs from zero for all values ofx . We solve the second-degree equation by completing the square,
i.e.
x+21
2−212−2x+212x+21=0
=49=
23x=−21−23=−2 andx=−21+23=1 . Both of these points lie within the region of definition,−3 .
x3 - The function is a polynomial
x2−x−1 multiplied by the exponential functionex , and, because both of these functions are differentiable, the product is also a differentiable function, which shows that the function is differentiable everywhere. - The function's region of definition is
−3 and the endpointsx3x=−3 andx=3 are therefore possible local extreme points.
All in all, there are four points
Now, we will write down a table of the sign of the derivative, in order to investigate if the function has local extreme points.
We can factorize the derivative somewhat,
| (x)=(x2+x−2)ex=(x+2)(x−1)ex |
since
| | | | | | |||
| | | | | | \displaystyle + | \displaystyle + | \displaystyle + |
| \displaystyle x-1 | \displaystyle - | \displaystyle - | \displaystyle - | \displaystyle - | \displaystyle 0 | \displaystyle + | \displaystyle + |
| \displaystyle e^x | \displaystyle + | \displaystyle + | \displaystyle + | \displaystyle + | \displaystyle + | \displaystyle + | \displaystyle + |
The sign of the derivative is the product of these signs and from the derivative's sign we decide which local extreme points we have:
| \displaystyle x | \displaystyle -3 | \displaystyle -2 | \displaystyle 1 | \displaystyle 3 | |||
| \displaystyle f^{\,\prime}(x) | \displaystyle + | \displaystyle 0 | \displaystyle - | \displaystyle 0 | \displaystyle + | ||
| \displaystyle f(x) | \displaystyle 11e^{-3} | \displaystyle \nearrow | \displaystyle 5e^{-2} | \displaystyle \searrow | \displaystyle -e | \displaystyle \nearrow | \displaystyle 5e^3 |
The function has local minimum points at \displaystyle x=-3 and \displaystyle x=1, and local maximum points \displaystyle x=-2 and \displaystyle x=3.
