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Magazine Chapter 10: Problem 42
Sketch the graph of the function, using the curve-sketching quide of this section. $$ f(t)=2 t^{3}-15 t^{2}+36 t-20 $$
Short Answer
Expert verified
The graph of the function \(f(t) = 2t^3 - 15t^2 + 36t - 20\) has a y-intercept at (0, -20), critical points at t = 2 (local maximum) and t = 3 (local minimum), a point of inflection at t = 2.5, concave up for t < 2.5, and concave down for t > 2.5. The graph approaches positive infinity as t approaches positive infinity and approaches negative infinity as t approaches negative infinity.
Step by step solution
01
Determine the intercepts
First, find the t-intercepts by setting f(t) equal to 0 and solving for t:
\(2t^3 - 15t^2 + 36t - 20 = 0\)
This equation cannot be easily factored. We can use numerical methods like the Rational Root Theorem to find the roots or use graphing technology, either way, we will skip this step and focus on the later steps of the curve-sketching guide.
Next, find the y-intercept by plugging t = 0 into the function:
\(f(0) = 2(0)^3 - 15(0)^2 + 36(0) - 20 = -20\)
So, the y-intercept is at the point (0, -20).
02
Determine the critical points
To find critical points, like local maxima and minima, we need to check where the derivative of f(t) equals 0 or is undefined.
First, find the derivative of the function f(t):
\(f'(t) = 6t^2 - 30t + 36\)
Now, set f'(t) equal to 0 and solve for t:
\(6t^2 - 30t + 36 = 0\)
Factor out the greatest common denominator to simplify:
\(6(t^2 - 5t + 6) = 0\)
Now, factor the quadratic within the parentheses:
\(6(t - 3)(t - 2) = 0\)
So, the critical points are at t = 2 and t = 3.
03
Determine the concavity and points of inflection
To analyze the concavity of the graph, we need to find the second derivative of f(t) and check its sign.
First, find the second derivative of f(t):
\(f''(t) = 12t - 30\)
Now, we find the point(s) of inflection by setting f''(t) equal to 0 and solving for t:
\(12t - 30 = 0\)
\(t = \frac{30}{12} = 2.5\)
Now, analyze the concavity by plugging in values of t before, within, and after this point of inflection into the second derivative.
For t < 2.5 (e.g., t = 1), f''(t) > 0, which means the graph is concave up.
For t > 2.5 (e.g., t = 3), f''(t) < 0, which means the graph is concave down.
So, the point of inflection occurs at t = 2.5.
04
Analyze the end behavior
As t approaches positive or negative infinity, we need to analyze the end behavior of the function.
Because f(t) is a cubic function with a positive leading coefficient, we know that as t approaches positive infinity, the function approaches positive infinity. As t approaches negative infinity, the function approaches negative infinity.
05
Sketch the graph
Using the information obtained from the previous steps, sketch the graph indicating the intercepts, critical points, concavity change, point of inflection, and end behavior.
1. Plot the y-intercept at (0, -20).
2. Mark the critical points at t = 2 and t = 3, and show a local maximum at t=2, and a local minimum at t=3.
3. Draw the graph as concave up for t < 2.5, and concave down for t > 2.5.
4. Indicate the point of inflection at t = 2.5.
5. Show the end behavior as the function approaches positive infinity when t approaches positive infinity, and the function approaches negative infinity when t approaches negative infinity.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Critical Points
Finding critical points is essential for understanding where a function like \( f(t) = 2t^3 - 15t^2 + 36t - 20 \) has peaks or valleys. These points indicate where the slope of the tangent is zero, meaning the function is neither increasing nor decreasing. To find these critical points for our function, derive it to find \( f'(t) = 6t^2 - 30t + 36 \). By setting the derivative \( f'(t) \) equal to zero, you solve for \( t \) to find critical points: \( 6(t-3)(t-2) = 0 \). Thus, \( t = 2 \) and \( t = 3 \) are critical points where the slope of the tangent to the curve is zero.
- At \( t = 2 \), there is a local maximum.
- At \( t = 3 \), there is a local minimum.
Concavity
Concavity tells us how the graph curves, whether upwards like a cup or downwards like a cap. It can be determined using the second derivative. For our function \( f(t) = 2t^3 - 15t^2 + 36t - 20 \), the second derivative is \( f''(t) = 12t - 30 \). By looking at the sign of \( f''(t) \):
- If \( f''(t) > 0 \), the function is concave up.
- If \( f''(t) < 0 \), the function is concave down.
Intercepts
Intercepts are points where the graph crosses the axes. For a function like \( f(t) = 2t^3 - 15t^2 + 36t - 20 \), intercepts help position the curve on the coordinate system and show where \( f(t) = 0 \).**Finding Intercepts:**- **T-Intercepts**: These are found by solving the equation \( 2t^3 - 15t^2 + 36t - 20 = 0 \). If exact solutions are not simple to find factored, numeric or graphing methods are often used.- **Y-Intercept**: Set \( t = 0 \) to find the y-intercept. Here, \( f(0) = -20 \), so the y-intercept is at \( (0, -20) \). Identifying where the graph crosses the axes is crucial for sketching it effectively.
Points of Inflection
Points of inflection occur where the concavity of a graph changes, like from concave up to concave down or vice versa. They are determined by setting the second derivative equal to zero and solving for \( t \). For \( f(t) = 2t^3 - 15t^2 + 36t - 20 \), we found that the second derivative is \( f''(t) = 12t - 30 \). Setting \( f''(t) = 0 \), we solve for the inflection point:- \( 12t - 30 = 0 \) gives \( t = 2.5 \).At \( t = 2.5 \), the curve changes from concave up to concave down, marking a transition in the graph's curvature. Recognizing and plotting these changes can provide important insight into the graph's overall shape and behavior.
End Behavior
End behavior describes how a function behaves as it approaches infinity or negative infinity. For cubic functions like \( f(t) = 2t^3 - 15t^2 + 36t - 20 \), which have the highest degree of three, the end behavior is dominated by the leading term \( 2t^3 \).
- As \( t \rightarrow \infty \), because the leading coefficient (2) is positive, \( f(t) \rightarrow \infty \).
- As \( t \rightarrow -\infty \), \( f(t) \rightarrow -\infty \), due to the degree being odd and the leading coefficient positive.
