The Shape of a Graph, Part II — Question 7

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Question 7

Let f(x)=x3−6x2+12x−5f(x) = x^3 - 6x^2 + 12x - 5

(a) Determine the intervals where the graph of ff is concave up and concave down.

(b) Find all inflection points.

Original worksheet page 1: question and worked solution for 4-6-007
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Question 7 - Solution

We are given f(x)=x3−6x2+12x−5f(x) = x^3 - 6x^2 + 12x - 5

First derivative

f′(x)=3x2−12x+12f'(x) = 3x^2 - 12x + 12

Second derivative

f″(x)=6x−12f''(x) = 6x - 12

(a) Concavity

Setting f″(x)=0f''(x) = 0 gives x=2.x = 2.

For x<2x < 2, the second derivative is negative, so the graph is concave down.

For x>2x > 2, the second derivative is positive, so the graph is concave up.

Concave down on (−∞,2)\boxed{(-\infty, 2)}

Concave up on (2,∞)\boxed{(2, \infty)}

(b) Inflection point

Because the concavity changes at x=2x = 2, an inflection point occurs there.

f(2)=3f(2) = 3

Inflection point at (2,3)\boxed{(2, 3)}

Graph of f(x)f(x)

See the diagram in the original worksheet below.

Original worksheet page 2: question and worked solution for 4-6-007

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