Let g be a function that is defined for all x, x ≠ 2, such that g(3) = 4 and the derivative of g is g′(x)=${x^2–16}/{x−2}$, with x ≠ 2.

Find all values of x where the graph of g has a critical value.

For each critical value, state whether the graph of g has a local maximum, local minimum or neither. You must justify your answers with a complete sentence.

On what intervals is the graph of g concave down? Justify your answer.

Write an equation for the tangent line to the graph of g at the point where x = 3.

Does this tangent line lie above or below the graph at this point? Jus

Question

Let g be a function that is defined for all x, x ≠ 2, such that g(3) = 4 and the derivative of g is g′(x)=${x^2–16}/{x−2}$, with x ≠ 2.

Find all values of x where the graph of g has a critical value. 

For each critical value, state whether the graph of g has a local maximum, local minimum or neither. You must justify your answers with a complete sentence. 

On what intervals is the graph of g concave down? Justify your answer. 

Write an equation for the tangent line to the graph of g at the point where x = 3. 

Does this tangent line lie above or below the graph at this point? Jus

holsteremission · Answer

$\bullet$ $g(x)$ will have critical values wherever $g'(x)=0$.

$\bullet$ Each critical point can be determined to be the site of a local maximum or minimum by considering the concavity of $g(x)$ around each critical point. This involves checking the sign of the second derivative $g''(x)$ in the vicinity of a critical point $x=c$.

$\bullet$ A description of the function's concavity over the intervals between inflection points can be made by checking the sign of $g''(x)$ over each interval.

$\bullet$ The line tangent to $g(x)$ at $x=3$ will have slope $g'(3)$. You also know that at $x=3$, the function takes on the value of $g(3)=4$. Use the point-slope formula to find the equation of the line itself.
$$y-g(3)=g'(3)(x-3)\implies y=\cdots$$
$\bullet$ Checking whether the tangent line lies above or below $g(x)$ could be done in several ways. The most immediate method might be to simply check the sign of the difference of $g(x)$ and the tangent line above to either side of their intersection. (e.g. what's the value of $g(2.9)-y(2.9)$? of $g(3.1)-y(3.1)$ ?)

satellite73 · Answer

$$ g′(x)=\frac{x^2–16}{x−2}, with x ≠ 2. $$

satellite73 · Answer

critical points are where the derivative is zero, i.e. where $x^2-16=0$ and where it is undefined, i.e. where $x-2=0$

satellite73 · Answer

to find concavity, you have to find $g''$ , the derivative of $\frac{x^2-16}{x-2}$

sbentleyaz · Answer

so for critical points would that mean x=2 and x=4?

sbentleyaz · Answer

@satellite73