(a)Use differentiation to find the minimum value of f(x)= 3x^2 - 5x + 10 and the value(s) of x for which this minimum occurs.

Mathematics

1 answer:

elena-14-01-66 [18.8K]2 years ago

3 0

Hi there!

Begin by differentiating f(x) using the power rule:

dy/dx = nxⁿ⁻¹

Therefore:

f(x) = 3x² - 5x + 10

f'(x) = 6x - 5

Set this equation equal to 0 to find the x-intercept:

0 = 6x - 5

5 = 6x

x = 5/6, which is where the graph goes from NEGATIVE to POSITIVE, so there is a MINIMUM at this value.

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Express this to single logarithm

zzz [600]

Answer: $\log_{2}\left(\frac{q^2\sqrt{m}}{n^3}\right)$

We have something in the form log(x/y) where x = q^2*sqrt(m) and y = n^3. The log is base 2.

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Explanation:

It seems strange how the first two logs you wrote are base 2, but the third one is not. I'll assume that you meant to say it's also base 2. Because base 2 is fundamental to computing, logs of this nature are often referred to as binary logarithms.

I'm going to use these three log rules, which apply to any base.

log(A) + log(B) = log(A*B)
log(A) - log(B) = log(A/B)
B*log(A) = log(A^B)

From there, we can then say the following:

$\frac{1}{2}\log_{2}\left(m\right)-3\log_{2}\left(n\right)+2\log_{2}\left(q\right)\\\\\log_{2}\left(m^{1/2}\right)-\log_{2}\left(n^3\right)+\log_{2}\left(q^2\right) \ \text{ .... use log rule 3}\\\\\log_{2}\left(\sqrt{m}\right)+\log_{2}\left(q^2\right)-\log_{2}\left(n^3\right)\\\\\log_{2}\left(\sqrt{m}*q^2\right)-\log_{2}\left(n^3\right) \ \text{ .... use log rule 1}\\\\\log_{2}\left(\frac{q^2\sqrt{m}}{n^3}\right) \ \text{ .... use log rule 2}$