\(B(t)\) is decreasing on \((-\infty , -5)\) and \(B(t)\) is decreasing on \((-5, 0.5)\). Therefore, \(B(t)\) is decreasing on \((-\infty , -5) \cup (-5, 0.5)\)
True
False
features
Complete Analysis
![[Picture]](fullAnalysis2-0cdbefd042552661e08241f3d6d7beea.svg)
Completely analyze
\[ B(f) = \frac {(f+3)(f-1)}{(f+5)(f-2)^2} \]
The natural domain is all real numbers except \(-5\) and \(2\), which make the denominator equal to \(0\). These are singularities of \(B\). Otherwise, \(B\) is continuous. There are no discontinuities.
Domain = \((-\infty , -5) \cup (-5, 2) \cup (2, \infty )\)
The graph will help us with the range, so we’ll wait on that.
\(B\) has zeros at \(\answer {-3}\) and \(\answer {1}\). These make the numerator equal to \(0\). These are represented on the graph with \((-3,0)\) and \((1,0)\)
The graph will have vertical asymptotes \(f=-5\) and \(f=-2\). \(B\) will change sign over \(f=-5\), since the multiplicity is odd even . \(B\) will not change sign over \(f=2\), since the multiplicity is odd even.
\(B\) is a rational function and the degree of the denominator is greater than less than the degree of the numerator. Therefore,
\[ \lim _{f \to -\infty } B(f) = \answer {0} \, \text { and } \, \lim _{f \to \infty } B(f) = \answer {0} \]
The graph has a horizontal asymptote at \(y=\answer {0}\).
From this, we can sketch a graph of \(y = B(f)\).
Both zeros have odd multiplicity, therefore the graph must go below the \(f\)-axis between them. There must be a local minimum. From the graph we can estimate that the critical number is approximately \(0.5\). \(B(0.5) = \frac {14}{99} \approx -0.1414\) is a local minimum. With some Calculus tools, we may be able to get an exact value.
\(B\) has no global maximum or minimum.
\(B\) has no local maximum.
- \(B\) decreases on \((-\infty , -5)\).
- \(B\) decreases on \((-5, 0.5)\).
- \(B\) increases on \((0.5, 2)\).
- \(B\) decreases on \((2, \infty )\).
The graph makes it evident that the range is all real numbers.
With some graphing tools, we can get a better approximation of the critical number.
The critical number is approximately \(0.181\) and the local minimum value of \(B\) is approximately \(-1.152\).
Analyze
![[Picture]](fullAnalysis2-0f80d10139f9d15fc81f31433044e88b.svg)
Completely analyze \(K(w) = w \sqrt {9-w}\).
The domain is restricted by the square root to be \(\left ( -\infty , \answer {9} \right ]\).
The formula is a product, therefore the zeros come from each factor.
The factor of \(w\) gives a zero of \(\answer {0}\).
The factor of \(\sqrt {9-w}\) gives a zero of \(\answer {9}\).
Since \(\sqrt {9-w}\) is always nonnegative, the sign of \(K\) is the same as the sign of \(w\).
- \(K\) is negative on \(\left ( \answer {-\infty },\answer {0} \right )\)
- \(K\) is positive on \(\left ( \answer {0}, \answer {\infty } \right )\)
From this, we can sketch a graph of \(y = K(w)\).
Our sketch suggest a global maximum between \(0\) and \(9\).
If we had the derivative, then we could identify this critical number exactly. Currently, we’ll need a graph to approximate values.
- \(K\) has a local and global maximum of \(10.39\) at \(6\).
- \(K\) has a local minimum of \(0\) at \(9\).
- \(K\) has no global minimum.
- \(K\) is increasing on \((-\infty , 6]\).
- \(K\) is decreasing on \([6, 9]\).
Finally, the range is approximately \((-\infty , 10.39]\).
The graph has no vertical asymptotes, \(\lim \limits _{w \to -\infty } K(w) = -\infty .\)
We do have alternatives to the derivative for some types of functions.
Critical Number
![[Picture]](fullAnalysis2-8b856d2c2c7c8f55bcf7e7c3968e1392.svg)
Earlier, we saw an algebraic method of identifying the critical number in the previous example.
We think there is a hump in the graph of \(K\) and at the highest point, \((A, A\sqrt {9-A})\), the tangent line would be horizontal.
Now, create a new function called \(F\).
\[ F(x) = K(x) - K(A) = K(x) - A\sqrt {9-A} \]
As we saw earlier, since we have a tangent line, the difference of the original function and the linear function has a double root at \(A\). \(x-A\) divides evenly into \(F\). And, the resulting function then also has \(A\) as a root.
First, factor out \(x-A\) from \(F\).
\[ x\sqrt {9-x} - A\sqrt {9-A} = \frac {x\sqrt {9-x} - A\sqrt {9-A}}{1} \cdot \frac {x\sqrt {9-x} + A\sqrt {9-A}}{\answer {x\sqrt {9-x} + A\sqrt {9-A}}} \]
\[ = \frac {x^2 (9-x) - A^2 (9-A)}{x\sqrt {9-x} + A\sqrt {9-A}} = \frac {9x^2 - x^3 - 9A^2 + A^3}{x\sqrt {9-x} + A\sqrt {9-A}} \]
\[ = \frac {9(x^2-A^2) - (x^3 - A^3)}{x\sqrt {9-x} + A\sqrt {9-A}} = \frac {(x-A)(9(x+A)-(x^2 + xA + A^2))}{x\sqrt {9-x} + A\sqrt {9-A}} \]
\(x-A\) divides into this evenly leaving
\[ \frac {9(x+A)-(x^2 + xA + A^2)}{x\sqrt {9-x} + A\sqrt {9-A}} \]
\(A\) is a root of this, which means \(A\) is a root of the numerator.
\[ 9(A+A)-(A^2 + A^2 + A^2) = 0 \]
\[ 18A - 3A^2 = 0 \]
\[ 3A \left (\answer {6-A} \right ) = 0 \]
The zero we are looking for is \(6\). The critical number is \(6\) and \(K(6) = 6 \sqrt {3}\).
- \(K\) has a local and global maximum of \(\answer {6\sqrt {3}}\) at \(6\).
- \(K\) has a local minimum of \(\answer {0}\) at \(9\).
- \(K\) has no global minimum.
- \(K\) is increasing on \((-\infty , 6]\).
- \(K\) is decreasing on \([6, 9]\).
Finally, the range is \((-\infty , 6\sqrt {3}]\).
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