1. ## 3 Tricky Calculus Problems. Help Please.

I can not get these three problems. I tried and tried but nothing makes sense. If someone could walk me through it and explain them to me I would be very grateful.

1)Let $p$ and $q$ be real numbers and let $f$ be the function defined by

$f(x)$= $1+2p(x-1)+(x-1)^2$, for x less than or equal to 1 and $qx+p$, for x greater than 1

(a) find the values of $q$, in terms of $p$, for which $f$ is continuous at $x=1$
(b) find the values of $p$ and $q$ for which $f$ is differentiable at $x=1$
(c) if $p$ and $q$ have the values determined in part (b), is second derivative $f$ a continuous function? Justify you answer.

6)Let $P(x)=x^4+ax^3+bx^2+cx+d$. The graph $y=P(x)$ is symmetric with respect to the Y-axis, has relative maximum at $(0,1)$, and has an absolute minimum at $(q, -3)$
a) determine the values $a, b, c, d$ and using these values write an expression for $P(x)$
b)Find all possible values of $q$

7)given the curve $x^2-xy+y^2=9$
(a) write a general expression for the slope of the curve.
(b) Find the coordinates of the points on the curve where the tangents are vertical.
(c) at the point $(0,3)$ find the change in the slope of the curve with respect to x.

Thank you everybody for all your help!

2. For number 7, i think the slope formula is, $\frac{-2x-y}{x+2y}$ for (a) and (c) is $\frac{1}{2}$. I can not get (b) or the other problems. Any help would be greatly appreciated.

Thanks Again.

3. 1)Let and be real numbers and let be the function defined by

= , for x less than or equal to 1 and , for x greater than 1

(a) find the values of , in terms of , for which is continuous at
Use the definition of continuity: the limits from both directions and the value of the function must be the same for the function to be continuous. Set
$f(1) = \lim_{x\to 1} (1+2p(x-1)+(x-1)^2) = \lim_{x\to 1}(qx+p)$
and get q in terms of p.

(b) find the values of and for which is differentiable at
differentiable requires that $\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$ is defined. Plug in x=1 and see what, if any, restrictions you need to apply to p to get it to converge.

(c) if and have the values determined in part (b), is second derivative a continuous function? Justify you answer.
Take the 2nd derivative. at x=1, you will have to use the definition of derivative. When you have done this, it will be clear whether the 2nd derivative is continuous or not.

6)Let . The graph is symmetric with respect to the Y-axis, has relative maximum at , and has an absolute minimum at
a) determine the values and using these values write an expression for
You have lots of information that you need to translate into equations.
The graph is symmetric with respect to the Y-axis,
P(x) = P(-x)
has relative maximum at
2 pieces of information: the graph passes through $(0,1) \implies P(0) = 1$ and there is a local maximum at $x=0 \implies P'(0) = 0$
absolute minimum at
the graph passes through (q,-3) and P'(q)=0.

i think the slope formula is, for (a)
I got $\frac{2x-y}{2y-x}$. Check your work and post again if you are stuck.

7...(b) Find the coordinates of the points on the curve where the tangents are vertical.
The vertical tangents are where the gradient is infinite. Set the denominator equal to 0 and solve this simultaneously with the equation for the curve (so that the points you find are actually on the graph).

4. ## thank you so much!

Thank you so much. You really did help me out of a bind there. Your explanations were concise and clear. Thanks again and if you celebrate thanksgiving, Happy Thanksgiving!