MHB Limit Definitions and Extreme Value Theorem Help Needed

[ardentmed]

Hey guys,

I have a couple more questions about this problem set I've been working on. I'm doubting some of my answers and I'd appreciate some help.

Question:

For 1a, I just took \lim_{{h}\to{0}} of the function using
[ f(x+h)-f(x) / h ]
and simplified.

Ultimately, this gave me -8/(2√(1-8x)) which equals:

f(x) = -4/(√1-8x)
As for 1b, I just used the slope form of the function and substituted 1a's answer into the function to get:

y-3 = [-4/(√1-8x)] * (x+1)


As for 2a, I used the intermediate value theorem, stating that if f is continuous on [a,b] then N is any number between f(b) and f(a_ and thus f(c) = N exists.

To prove this f -> infinity as x-> 0+
and f -> infinity as x-> infinity proves that one real root must exist by the IVT. Is this right? I'm somewhat doubtful of my solution.

As for 2b, I just found the intercepting points to be [1.34,1.35] Did I estimate this properly? Thanks in advance.

 

[mathmari]

Hey guys,

I have a couple more questions about this problem set I've been working on. I'm doubting some of my answers and I'd appreciate some help.

Question:For 1a, I just took \lim_{{h}\to{0}} of the function using
[ f(x+h)-f(x) / h ]
and simplified.

Ultimately, this gave me -8/(2√(1-8x)) which equals:

f(x) = -4/(√1-8x)
As for 1b, I just used the slope form of the function and substituted 1a's answer into the function to get:

y-3 = [-4/(√1-8x)] * (x+1)


As for 2a, I used the intermediate value theorem, stating that if f is continuous on [a,b] then N is any number between f(b) and f(a_ and thus f(c) = N exists.

To prove this f -> infinity as x-> 0+
and f -> infinity as x-> infinity proves that one real root must exist by the IVT. Is this right? I'm somewhat doubtful of my solution.

As for 2b, I just found the intercepting points to be [1.34,1.35] Did I estimate this properly? Thanks in advance.

1a. The derivative is correct!

1b. The formula of the equation of the tangent line to $f$ at $x=x_0$ is
$$y-f(x_0)=f'(x_0)(x-x_0)$$2a. To use the intermediate value theorem you have to find a closed interval $[a,b]$. The interval that you used is not closed.

Using the interval that you chose, you could do the following:
$f=\ln{x}+2x-3$
The domain of $f$ is $(0, +\infty)$.
$f$ is increasing on this interval.
So the range of the function is $\displaystyle{R= \left ( \lim_{x \rightarrow 0} f(x), \lim_{x \rightarrow +\infty} f(x) \right )=(-\infty, +\infty)=\mathbb{R}}$

So, since $0 \in R$, the function has at least one root.To use the intermediate value theorem, you could use the interval $[1,2]$.

2b. It is correct!

 

[ardentmed]

1a. The derivative is correct!

1b. The formula of the equation of the tangent line to $f$ at $x=x_0$ is
$$y-f(x_0)=f'(x_0)(x-x_0)$$2a. To use the intermediate value theorem you have to find a closed interval $[a,b]$. The interval that you used is not closed.

Using the interval that you chose, you could do the following:
$f=\ln{x}+2x-3$
The domain of $f$ is $(0, +\infty)$.
$f$ is increasing on this interval.
So the range of the function is $\displaystyle{R= \left ( \lim_{x \rightarrow 0} f(x), \lim_{x \rightarrow +\infty} f(x) \right )=(-\infty, +\infty)=\mathbb{R}}$

So, since $0 \in R$, the function has at least one root.To use the intermediate value theorem, you could use the interval $[1,2]$.

2b. It is correct!

Thanks for the insightful response. Could you please clarify what you're referring to for 1b?

 

[Dethrone]

$$y-f(x_0)=f'(x_0)(x-x_0)$$ is just a fancy way of writing $$y=mx+b$$, where b is $f(x_0)$, and m is the derivative of the tangent or $f'(x_0)$

 

[mathmari]

Thanks for the insightful response. Could you please clarify what you're referring to for 1b?

When you calculated the tangent line at $x=-1$, you used the formula: y-f(-1)=f'(x)(x-(-1))

but it should be:
y-f(-1)=f'(-1)(x-(-1))