Let f be the function given by f(x)=4x2−x3, and let t be the line y=18−3x, where ℓ is tangent to the graph ∝+f. Let R be the region bounded by the graph of f and the x-axis, and let S be the region bounded by the graph of f, the line ℓ, and the x-axis, as shown.(a) Show that ℓ is tangent to the graph of f(x)=4x2−x34 at the point f(x)=4x2−x35.(b) Find the area of S.(c) Find the volume of the solid generated when R is revolved about the x-axis.
Q. Let f be the function given by f(x)=4x2−x3, and let t be the line y=18−3x, where ℓ is tangent to the graph ∝+f. Let R be the region bounded by the graph of f and the x-axis, and let S be the region bounded by the graph of f, the line ℓ, and the x-axis, as shown.(a) Show that ℓ is tangent to the graph of f(x)=4x2−x34 at the point f(x)=4x2−x35.(b) Find the area of S.(c) Find the volume of the solid generated when R is revolved about the x-axis.
Verify Tangent Line: Verify that the line y=18−3x is tangent to f(x) at x=3.To show that ℓ is tangent to f at x=3, we need to check two things: the value of f(3) and f′(3) (the derivative of f at x=3).Calculate f(x)0.Calculate the derivative f(x)1, then f(x)2.The equation of the tangent line at x=3 using point-slope form is f(x)4, simplifying to f(x)5.Since y=18−3x matches this, ℓ is indeed tangent to f at x=3.
Calculate Area of Region: Find the area of region S.Region S is bounded by f(x), ℓ, and the x-axis from x=0 to x=3.Calculate the area under f(x) from x=0 to x=3 using integration: ∫03(4x2−x3)dx.This integral evaluates to S0.Calculate the area under ℓ from x=0 to x=3: S4.This integral evaluates to S5.Area of S is S7 square units.
Find Volume of Solid: Find the volume of the solid generated when R is revolved about the x-axis.Use the disk method to find the volume of the solid formed by revolving R around the x-axis from x=0 to x=3.Volume V=π∫03(4x2−x3)2dx.This integral is complex, but simplifies to π[34x3−41x4]03 squared.Evaluating this integral incorrectly leads to a wrong volume calculation.
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