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3 years ago

Given the function f(x)=x2+2x+1, find: f( (2a−3)/5 ) = Show working please!

Mathematics

1 answer:

Rina8888 [55]3 years ago

7 0

<h3>Given</h3>

f(x) = x² + 2x + 1

f((2a-3)/5)

<h3>Solution</h3>

It can be convenient to rewrite f(x) as a square, then do the substitution. That way, the algebra is simplified a little bit.

... f(x) = (x +1)²

... f((2a-3)/5) = ((2a-3)/5 +1)² = ((2a -3 +5)/5)²

... = (2/5(a+1))²

... f((2a-3)/5) = (4/25)(a² +2a +1)

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Find the limit , picture provided

V125BC [204]

Answer:

d. does not exist

Step-by-step explanation:

The given limits are;

$\lim_{x \to 4} f(x) =5$ , $\lim_{x \to 4} g(x) =0$ and $\lim_{x \to 4} h(x) =-2$

We want to find

$\lim_{x \to 4} \frac{f}{g}(x)= \lim_{x \to 4} \frac{f(x)}{g(x)}$

By the properties of limits, we have;

$\lim_{x \to 4} \frac{f}{g}(x)= \frac{\lim_{x \to 4} f(x)}{\lim_{x \to 4} g(x)}$

This gives us;

$\lim_{x \to 4} \frac{f}{g}(x)= \frac{5}{0}$

Division by zero is not possible. Therefore the limit does not exist.

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