# Thread: find a and b to make continuous

1. ## find a and b to make continuous

1) f(x) = ax^4 if |x| >= 1
(x^2) + b if |x|<1

I need to find a and b to make f continuous and differentiable everywhere.

I have no idea how though?

2) a)I need to find the domain and range of the derivative of h(x) = x + sqrt[x] which i found to be 1 + .5x^(-.5).
I was never good at finding domain and range.

b) determine, in standard form, the equation of the line normal to the curve, y = h(x), at point (1, h(1)).
(in case anyone doesnt know, normal means perpendicular to the tangent line)

1) f(x) = ax^4 if |x| >= 1
(x^2) + b if |x|<1

I need to find a and b to make f continuous and differentiable everywhere.

I have no idea how though?
what does the definition of continuity say? you have to fulfill three conditions.

a function is continuous at a point if:
(1) the function is defined at the point
(2) the limit as x goes to the point exists
(3) the limit is equal to the value of the function

as for differentiable everywhere. note that a function $f(x)$ is differentiable at a point $x$ if the limit $\lim_{h \to 0} \frac {f(x + h) - f(x)}h$ exists. we then denote this limit by $f'(x)$

Hint: if the function is defined piecewise about the point $x$, we need to ensure $\lim_{h \to 0^-} \frac {f(x + h) - f(x)}h = \lim_{h \to 0^+} \frac {f(x + h) - f(x)}h$

or, just find the derivative of the pieces and make sure it is continuous in much the same way as you made the original function continuous.

2) a)I need to find the domain and range of the derivative of h(x) = x + sqrt[x] which i found to be 1 + .5x^(-.5).
I was never good at finding domain and range.
do you know what the domain and range are?

b) determine, in standard form, the equation of the line normal to the curve, y = h(x), at point (1, h(1)).
(in case anyone doesnt know, normal means perpendicular to the tangent line)
you are looking for a creature that looks like $y = mx + b$, this is what we call a straight line. $m$ is the slope, and $b$ is the y-intercept. you should be aware, that the derivative gives the formula for the slope at any value of x. thus, the slope of the tangent line at the point where x = 1 is $h'(1)$. therefore, the slope of the normal line is $\frac {-1}{h'(1)}$.

thus, you are looking for a line with slope $m = \frac {-1}{h'(1)}$ passing through the point $(1, h(1))$. i suppose you can find the equation of a line given this information.

the standard form of a line is $ax + by + c = 0$ where $a,~b, \text{ and }c$ are constants

3. ok. 2b i found out.

2a i found the domain as (0, infinity) but how do i find the range?

and #1 i still dont understand. How do I even start it?

ok. 2b i found out.
good

2a i found the domain as (0, infinity)
nice!

but how do i find the range?
the range is the corresponding y-values for the domain. so to find the range, figure out the possible values that y can be when x is in $(0, \infty)$. when x is close to zero, what does the y-value get close to? what about when x goes to infinity?

and #1 i still dont understand. How do I even start it?
to start you off:

note that the function breaks into pieces around the points x = -1 and x = 1

so for starters, you need to make sure $\lim_{x \to 1^-}f(x) = \lim_{x \to 1^+} f(x)$ and $\lim_{x \to -1^-}f(x) = \lim_{x \to -1^+} f(x)$
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