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Question: How do you test \( f\left( x \right) = 8{x^4} - 9{x^3} + 9 \) for concavity and inflection points?...

How do you test f(x)=8x4−9x3+9f\left( x \right) = 8{x^4} - 9{x^3} + 9 for concavity and inflection points?

Explanation

Solution

Hint : In order to test for concavity and inflection points for the given function, first find the second order derivative of the function, then equate it to zero, if for any no. it gives zero that is its inflection point then simply check for the concave part by taking two values one less than the inflection point and another greater than that in second order derivative.
Formula used:
dxndx=nxn−1\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}

Complete step-by-step answer :
We are given the function f(x)=8x4−9x3+9f\left( x \right) = 8{x^4} - 9{x^3} + 9 .
We need to find the second order derivative of the function and to reach there, first derivate it for first order derivative with respect to xx and we get:
f(x)=8x4−9x3+9 f′(x)=32x3−27x2   f\left( x \right) = 8{x^4} - 9{x^3} + 9 \\\ f'\left( x \right) = 32{x^3} - 27{x^2} \;
After this moving ahead for our second derivative and we get:
f(x)=8x4−9x3+9 f′(x)=32x3−27x2 f′′(x)=96x2−54x   f\left( x \right) = 8{x^4} - 9{x^3} + 9 \\\ f'\left( x \right) = 32{x^3} - 27{x^2} \\\ f''\left( x \right) = 96{x^2} - 54x \;
Since, we are not getting zero from the second order derivative, so equating it with zero to obtain the value of xx and we get:
f′′(x)=96x2−54x=0f''\left( x \right) = 96{x^2} - 54x = 0
Taking 6x6x common from the above equation and solving further:
96x2−54x=0 6x(16x−9)=0 x=0,x=916   96{x^2} - 54x = 0 \\\ 6x\left( {16x - 9} \right) = 0 \\\ x = 0,x = \dfrac{9}{{16}} \;
Therefore, the inflection points for the function are x=0x = 0 and x=916x = \dfrac{9}{{16}} .
Choosing auxiliary value −1- 1 (as it is less than x=0x = 0 ) and substituting it in f′′(x)f''\left( x \right) to check for concavity and we get:
f′′(−1)=96(−1)2−54(−1)=96+54=150f''\left( { - 1} \right) = 96{\left( { - 1} \right)^2} - 54\left( { - 1} \right) = 96 + 54 = 150 that is 150>0150 > 0 which means at (∞,0)\left( {\infty ,0} \right) , the curve concaves upwards (as positive values concaves upwards).
Choosing auxiliary value 12\dfrac{1}{2} , as 12\dfrac{1}{2} is less than 916\dfrac{9}{{16}} and greater than 00 , substituting the auxiliary value in f′′(x)f''\left( x \right) , and we get:
f′′(12)=96(12)2−54×12=24−27=−3f''\left( {\dfrac{1}{2}} \right) = 96{\left( {\dfrac{1}{2}} \right)^2} - 54 \times \dfrac{1}{2} = 24 - 27 = - 3 , that is −3<0- 3 < 0 which means at (0,916)\left( {0,\dfrac{9}{{16}}} \right) , the curve concave downward (as negative value concaves downwards).
And lastly, substituting 11 in f′′(x)f''\left( x \right) as 11 is greater than 916\dfrac{9}{{16}} , to check for concavity and we get:
f′′(1)=96(1)2−54×1=96−54=42f''\left( 1 \right) = 96{\left( 1 \right)^2} - 54 \times 1 = 96 - 54 = 42 , that is 42>042 > 0 which means at (916,∞)\left( {\dfrac{9}{{16}},\infty } \right) , the curve concaves upwards.
Thus, for the function f(x)=8x4−9x3+9f\left( x \right) = 8{x^4} - 9{x^3} + 9 , at (∞,0)\left( {\infty ,0} \right) and (916,∞)\left( {\dfrac{9}{{16}},\infty } \right) concavity is upwards.
And, at (0,916)\left( {0,\dfrac{9}{{16}}} \right) concavity is downwards.
And, this is how we check for concavity and inflection points.

Note : If the second order derivative comes constant and if its greater than zero that means the curve will always remain concave upward and if it is less than zero then it will remain concave downwards.
There can be errors with the wrong formula of derivatives.