f(0)=0, so the y-intercept is 0. Solve x^3-3x^2=0 x^2(x-3)=0, giving x-intercepts at x=0 and x=3.

f''(x)=6x-6. - f''(0)=-6 0, so (2,-4) is a local minimum.

SAMath MethodsQuick questions

Topic 1: Further Differentiation and Applications

Quick questions on Curve sketching with derivatives (SACE Stage 2 Mathematical Methods)

Q: What are stationary points?

f'(x)=3x^2-6x=3x(x-2). Setting f'(x)=0 gives x=0 and x=2. - f(0)=0(0,0).

Q: What is inflection?

f''(x)=6x-6=0 at x=1. Sign changes from negative to positive, so (1,f(1))=(1,-2) is a point of inflection.

Q: What is end behaviour?

As x→∞, f(x)→∞; as x→-∞, f(x)→-∞ (cubic with positive leading coefficient).

5short Q&A pairs drawn directly from our worked dot-point answer. For full context and worked exam questions, read the parent dot-point page.

What are intercepts?

Show answer

f(0)=0

, so the

y

-intercept is

0

. Solve

x^3-3x^2=0\Rightarrow x^2(x-3)=0

, giving

x

-intercepts at

x=0

and

x=3

What are stationary points?

Show answer

f'(x)=3x^2-6x=3x(x-2)

. Setting

f'(x)=0

gives

x=0

and

x=2

. -

f(0)=0\Rightarrow(0,0)

What is classify?

Show answer

f''(x)=6x-6

. -

f''(0)=-6<0

, so

(0,0)

is a local maximum. -

f''(2)=6>0

, so

(2,-4)

is a local minimum.

What is inflection?

Show answer

f''(x)=6x-6=0

x=1

. Sign changes from negative to positive, so

(1,f(1))=(1,-2)

is a point of inflection.

What is end behaviour?

Show answer

x\to\infty

f(x)\to\infty

; as

x\to-\infty

f(x)\to-\infty

(cubic with positive leading coefficient).

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