Question

Please solve this question no C...​

Answer 1

Answer:

$\frac{2}{3a^{\frac{1}{3}}}$

Step-by-step explanation:

$\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a}=f'(a)$.

$f(x)=x^\frac{2}{3}$

Using power rule we get that $f'(x)=\frac{2}{3}x^{-\frac{1}{3}}$

Evaluating this at $x=a$ gives us: $f'(a)=\frac{2}{3}a^{-\frac{1}{3}}$.

We could write without negative exponent giving us:

$f'(a)=\frac{2}{3a^{\frac{1}{3}}}$.

We could also go about it an algebraic way.

Notice the numerator is a difference of squares and can be factored as $(x^\frac{1}{3}-a^\frac{1}{3})(x^\frac{1}{3}+a^\frac{1}{3})$.

We need a factor in the numerator to be $x-a$ so we can get rid of the $x-a$ on bottom and then substitute $a$ for $x$.

Recall the difference of cubes formula:

$p^3-q^3=(p-q)(p^2+pq+q^2)$

We are going to use this on the denominator:

$(x^\frac{1}{3})^3-(a^\frac{1}{3})^3$

$(x^\frac{1}{3}-a^\frac{1}{3})(x^\frac{2}{3}+x^\frac{1}{3}a^\frac{1}{3}+a^\frac{2}{3})$

So that first factor there will actually cancel with a factor I mentioned for the numerator earlier.

Let's see it all together:

$\lim_{x \rightarrow a}\frac{(x^\frac{1}{3}-a^\frac{1}{3})(x^\frac{1}{3}+a^\frac{1}{3})}{(x^\frac{1}{3}-a^\frac{1}{3})(x^\frac{2}{3}+x^\frac{1}{3}a^\frac{1}{3}+a^\frac{2}{3})}$

After the cancellation we have:

$\lim_{x \rightarrow a}\frac{(x^\frac{1}{3}+a^\frac{1}{3})}{(x^\frac{2}{3}+x^\frac{1}{3}a^\frac{1}{3}+a^\frac{2}{3})}$

Now we are ready to replace $x$ with $a$.

$\frac{a^\frac{1}{3}+a^\frac{1}{3}}{a^\frac{2}{3}+a^\frac{1}{3}a^\frac{1}{3}+a^\frac{2}{3}}$

We have some like terms to combine:

$\frac{2a^{\frac{1}{3}}}{a^\frac{2}{3}+a^{\frac{2}{3}}+a^\frac{2}{3}}$

$\frac{2a^\frac{1}{3}}{3a^\frac{2}{3}}$

$\frac{2}{3a^{\frac{2}{3}-\frac{1}{3}}}$

$\frac{2}{3a^{\frac{1}{3}}}$