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2 years ago

A function has f''(x) = 10 and has f'(4) = 0 and f(2) = 4. Find f(x)

Mathematics

1 answer:

labwork [276]2 years ago

3 0

Answer:

f(x) = 5x^2 - 40x + 64

Step-by-step explanation:

f''(x) = 10

f'(x) = 10x + a

f'(4) = 0 = 10*4 + a

a = -40

So f'(x) = 10x - 40

f(x) = 5x^2 - 40x + b

f(2) = 4 = 5*2^2 - 40*2 + b

4 = 20 - 80 + b

b = 64

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irakobra [83]

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2 years ago

Let f(x)=7x-13. Find f^-1(x).

vitfil [10]

Ok but allow my humble self to use $y$ instead of $f(x)$ .

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Hope this helps.

5 0

3 years ago

Read 2 more answers

3^x= 3*2^x solve this equation

kompoz [17]

In the equation

$3^x = 3\cdot 2^x$

divide both sides by $2^x$ to get

$\dfrac{3^x}{2^x} = 3 \cdot \dfrac{2^x}{2^x} \\\\ \implies \left(\dfrac32\right)^x = 3$

Take the base-3/2 logarithm of both sides:

$\log_{3/2}\left(\dfrac32\right)^x = \log_{3/2}(3) \\\\ \implies x \log_{3/2}\left(\dfrac 32\right) = \log_{3/2}(3) \\\\ \implies \boxed{x = \log_{3/2}(3)}$

Alternatively, you can divide both sides by $3^x$ :

$\dfrac{3^x}{3^x} = \dfrac{3\cdot 2^x}{3^x} \\\\ \implies 1 = 3 \cdot\left(\dfrac23\right)^x \\\\ \implies \left(\dfrac23\right)^x = \dfrac13$

Then take the base-2/3 logarith of both sides to get

$\log_{2/3}\left(2/3\right)^x = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x \log_{2/3}\left(\dfrac23\right) = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x = \log_{2/3}\left(3^{-1}\right) \\\\ \implies \boxed{x = -\log_{2/3}(3)}$