# calculus

• December 20th 2008, 10:53 AM
calculus
let f be a function defined by f(x)=
2x- x^2 for x is less than or equal to 1
x^2 +kx +p for x>1

for what values of k and p will f be continuous and differentiable at x=1

• December 20th 2008, 11:19 AM
earboth
Quote:

let f be a function defined by f(x)=
2x- x^2 for x is less than or equal to 1
x^2 +kx +p for x>1

for what values of k and p will f be continuous and differentiable at x=1

You've got:

$f(x)=-x^2+2x~and~f'(x)=-2x+2$
and
$g(x)=x^2+kx+p$

You know f(1)=1 that means the endpoint of f is P(1,1) and the slope of the graph at P is zero.

$\lim_{x\rightarrow1}(x^2+kx+p)=1~\implies~k = -p$

$\lim_{x\rightarrow1}(2x+k)=0~\implies~k = -2$

Therefore p = 2

The equation of g: $g(x) = x^2-2x+2$
• December 20th 2008, 11:21 AM
running-gag
Hi

The continuity at x=1 is expressed by
$\lim_{x \rightarrow 1^+} f(x) = f(1)$

$\lim_{x \rightarrow 1^+} (x^2 + kx + p) = 1$

$1 + k + p = 1$

$k + p = 0$

The differentiability at x=1 is expressed by
$\lim_{x \rightarrow 1^+} f'(x) = f'(1)$

$\lim_{x \rightarrow 1^+} (2x + k) = 0$

$2 + k = 0$

Then $k = -2$ and $p = 2$