Grade 12 · Cubic derivatives · 387 of 600
Cubic derivatives · Cubic coefficient a=0; Linear coefficient b=1.5
Evaluate at x=0.4. Find the function value and first and second derivatives at that input. Givens: Cubic coefficient a=0; Linear coefficient b=1.5.
Try the question, use a hint when you need one, and compare your reasoning with the worked solution. The animation starts with this chapter’s values; changing its controls explores a new case.
01 · Make a prediction
Your practice question
Let f(x)=(0)x³+(1.5)x. Evaluate at x=0.4. Find the function value and first and second derivatives at that input.
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A starting point
Differentiate each monomial: f′=3ax²+b and f″=6ax. A zero cubic coefficient gives a linear special case.
Work through the reasoning
Step 1
Identify the model and target
Let f(x)=(0)x³+(1.5)x. Evaluate at x=0.4. The governing relation is f(x)=ax³+bx; f′=3ax²+b; f″=6ax. Differentiate each monomial: f′=3ax²+b and f″=6ax. A zero cubic coefficient gives a linear special case.
Step 2
Substitute and calculate
At x=0.4, f=(0)(0.4)³+(1.5)(0.4)=0.6. Differentiation gives f′=3(0)(0.4)²+(1.5)=1.5 and f″=6(0)(0.4)=0.
Step 3
Check the mathematical meaning
At x=0.4, f′=1.5 and f″=2.4(0)=0. The second derivative's sign determines local upward or downward curvature when it is nonzero. Input x: 0.4; Function value: 0.6; First derivative: 1.5; Second derivative: 0. Decimal values are rounded, so use unrounded intermediate values.
The answer
At x=0.4, f=(0)(0.4)³+(1.5)(0.4)=0.6. Differentiation gives f′=3(0)(0.4)²+(1.5)=1.5 and f″=6(0)(0.4)=0. At x=0.4, f′=1.5 and f″=2.4(0)=0. The second derivative's sign determines local upward or downward curvature when it is nonzero. Animation check: Input x: 0.4; Function value: 0.6; First derivative: 1.5; Second derivative: 0. Decimal displays are rounded; retain the original parameters and exact π until the final step.
Compare the method as well as the result. A different valid method may reach the same answer. Keep exact values until the last step when the question asks for rounding.
02 · Explore the model
See the mathematical relationship move.
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Watch the relationship
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From experiment to screen.
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The mathematical idea
A cubic's height, slope, and change of slope answer different questions. The moving tangent shows why a point with zero height need not have zero slope. Varying the linear term can add or remove stationary points while the second derivative still changes sign at zero when a is nonzero. Let f(x)=(0)x³+(1.5)x. Evaluate at x=0.4. The requested state occurs at 60% playback; use the exact target stated in the question for your calculation.
f(x)=ax³+bx; f′=3ax²+b; f″=6ax
03 · Reflect and transfer
Explain what changes and why.
Change the linear coefficient alone. Which derivative changes and which curvature calculation remains the same?
This is a distinct guided scenario using a reusable mathematical model. Similar-looking diagrams can represent different given values and conclusions; they are not different mathematical theories.