# 3 problems from Calculus

• Feb 26th 2009, 02:22 AM
Mathhelpz
3 problems from Calculus
I'm been having some problems with these and I really need the help

1) The number of inflection points for the graph of

y=2x+cos(x^2) in the interval 0≤x≤5 is

2) If f(x) = 2x^2 -x^3 and g(x) = x^2-2x, for what values of a and b is

the integral from a to b f(x) dx > the integral from a to b g(x) dx?

The answer is a=0 and b=2

3)Also what does the lim as h approaches 0 when f(2+h)-f(2)/h=2 look like? Or rather what does it do.
• Feb 26th 2009, 08:14 AM
Soroban
Hello, Mathhelpz!

These are not simple problems.

Quote:

1) The number of inflection points for the graph of: .$\displaystyle y\:=\:2x+\cos(x^2)$
in the interval $\displaystyle 0\leq x \leq 5$

We must solve: .$\displaystyle y'' \:=\:0$

We have: .$\displaystyle y' \;=\;2-2x\sin(x^2)$

Then: .$\displaystyle y'' \;=\;-4x^2\cos(x^2) - 2\sin(x^2) \:=\:0$

. . $\displaystyle 2\sin(x^2) \:=\:-4x^2\cos(x^2) \quad\Rightarrow\quad \frac{\sin(x^2)}{\cos(x^2)} \:=\:-2x^2$

And we must solve: .$\displaystyle \tan(x^2) \:=\:-2x^2$

I recommend a graphing calculator to find the intersections of:
. . $\displaystyle y \:=\:\tan(x^2)\,\text{ and }\,y \:=\:-2x^2$

Quote:

If $\displaystyle f(x) \:= \:2x^2 -x^3$ and $\displaystyle g(x) \:=\: x^2-2x$

for what values of a and b is: .$\displaystyle \int^b_af(x)\,dx \;> \; \int ^b_ag(x)\,dx$ ?

The answer is: $\displaystyle a=0, b=2$ . I don't agree

We have: .$\displaystyle f(x) \:=\:x^2(2-x)$
It is a "negative" cubic with x-intercepts 0 and 2.
. . The graph is tangent to the x-axis at $\displaystyle x = 0.$

We have: .$\displaystyle g(x) \:=\:x(x-2)$
It is an up-opening parabola with x-intercepts 0 and 2.

Those integrals are comparing the areas under the curves.
The question becomes: .When is the cubic above the parabola?
. .
I tried to sketch the graphs, but failed abysmally.

The curves intersect at $\displaystyle x \:=\: \text{-}1, 0, 2$

The cubic is above the parabola on the interval $\displaystyle 0 < x < 2$

But it is also above the parabola for: $\displaystyle x < \text{-}1$

Quote:

3) What does: .$\displaystyle \lim_{h\to0}\frac{f(2+h)-f(2)}{h}\;=\;2$ look like?

Or rather what does it do?

The expression: .$\displaystyle \lim_{h\to0}\frac{f(2+h) - f(2)}{h}\;=\;2$

. . says: the tangent to the graph of $\displaystyle y \,=\,f(x)$ at $\displaystyle (2,f(2))$ has slope 2.

The graph might look like this:
Code:

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