# Thread: Curve Sketching - Concavity - Problem 1

1. ## Curve Sketching - Concavity - Problem 1

Would anyone be so kind as to help me solving this concavity problem?
I just don't get concavity at all, and need to have a solution for this problem. Thank you very much in advance.

2. Try not to do these problems without 'getting' concavity - it's one of the easier things to 'get', I think, and basic to much else.

Imagine you're a bug driving a tiny car along the line of a graph on a page, from left to right.

Are you heading 'up' (y increasing, i.e. towards the top of the page)? Or down? You decide. (Imagine.)

But is your steering wheel turned to the right? If so, you're concave down. If you're heading up, you'll be levelling off soon and then coming down. If you're already heading down then things are getting dangerous!

But if your steering wheel is turned to the left, you're concave up. (Fill in the details...)

Differentiating the curve and plugging in an x-value tells you whether you're headed up or down. Differentiating again tells you whether your steering wheel is left (concave up) or right (concave down).

Solving

first derivative = 0

tells you at what x value(s) the graph is heading neither up nor down (is momentarily level).

Soving

second derivative = 0

tells you at what x value(s) your steering wheel is dead center. If this is momentarily between being turned left (concave up) and right (concave down) then this point is called a point of inflection...

So differentiate your function to investigate where it's going up, down, turning and inflecting...

3. $\displaystyle f(x)=-2x^4-7x^3+7$

$\displaystyle f'(x)=-8x^3-21x^2$

$\displaystyle f''(x)=-24x^2-42x$

$\displaystyle f''(x)=0$ at a point of inflection.

$\displaystyle f''(x)=0$ at $\displaystyle x=0$ and $\displaystyle x=-42/24=-1.75$

f has inflection points when $\displaystyle x=-1.75$ and $\displaystyle x=0$.

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The function is concave down if $\displaystyle f''(x)<0$.

$\displaystyle -24x^2-42x<0$

$\displaystyle x(-24x-42)<0$

$\displaystyle x(24x+42)>0$

i) $\displaystyle x>0$ and $\displaystyle 24x+42>0$

ie $\displaystyle x>0$ and $\displaystyle x>-1.75$

ie $\displaystyle x>0$

ii) $\displaystyle x<0$ and $\displaystyle 24x+42<0$

ie $\displaystyle x<0$ and $\displaystyle x<-1.75$

ie $\displaystyle x<-1.75$

Hence, f is concave down on $\displaystyle (- \infty, -1.75)$ U $\displaystyle (0, \infty)$

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The function is concave upward if $\displaystyle f''(x)>0$

$\displaystyle -24x^2-42x>0$

$\displaystyle x(-24x-42)>0$

$\displaystyle x(24x+42)<0$

i) $\displaystyle x>0$ and $\displaystyle 24x+42<0$

ie $\displaystyle x>0$ and $\displaystyle x<-1.75$

There is no such interval.

ii) $\displaystyle x<0$ and $\displaystyle 24x+42>0$

ie $\displaystyle x<0$ and $\displaystyle x>-1.75$

Hence, f is concave up on $\displaystyle (-1.75, 0)$