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The graph is concave up when the tangent line is below the graph and concave down is the graph where tangent line is above the graph. This is clear from the diagram below. To find whether the graph is concave upward or concave downward we will use the 2nd derivative test.

Let 'f' be a function whose 2nd derivative exists on an open interval I.

When the function y = f(x) is

When the function y = f(x) is

To find the 1st derivative we will use the quotient rule

f '(x) = $f'(x)= \frac{(x+1).\frac{\text{d}}{\text{d}y}(3x)- 3x.\frac{\text{d}}{\text{d}y}(x+1)}{(x+1)^{2}}$

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

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

f '(x) = $\frac{3 }{(x+1)^{2}}$

Now we will find the critical points

$(x+1)^{2}$ = 0

x = -1

So the interval will be ($-\infty$, -1) (-1,$\infty$)

To find the interval of concavity we will find the 2nd derivative

f '(x) = $\frac{3 }{(x+1)^{2}}$

f "(x) = $\frac{(x+1)^{2}.\frac{\text{d}}{\text{d}y}(3)- 3.\frac{\text{d}}{\text{d}y}(x+1)^{2}}{(x+1)^{4}}$

= $\frac{-3.2(x+1).1}{(x+1)^{4}}|$ -----(by using the chain rule)

f "(x) = $ \frac{-6(x+1)}{(x+1)^{4}}$
Now we will consider the any point from the above interval to check the concave up or down

Let us consider x = -2

f "(x) = $ \frac{-6(x+1)}{(x+1)^{4}}$

f "(-2) = $ \frac{-6(-2+1)}{(-2+1)^{4}}$

f "(-2) = 6 > 0 **so the graph will be concave upward in the interval ($-\infty$, -1) **

Let us consider x = 1

f "(x) = $ \frac{-6(x+1)}{(x+1)^{4}}$

f "(1) = $ \frac{-6(1+1)}{(1+1)^{4}}$

f "(1) = $ \frac{-12}{16}$ < 0

1) Find the 1st derivative .

2) Set the 1st derivative equal to zero and then find the critical points. If derivative is in the fraction, then set the numerator equal to zero and denominator equal to zero. i

3) Set the critical points on the number line and then write the interval.

4) For concavity, find the 2nd derivative.

5) Plug in any point from the interval in the 2nd derivative and check it positive or negative and then write the interval of concavity.

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