FIND the point of inflexion of the function y=1/3 x³ - 5/2x² + 6x - 12
Answers
Answer:
f ′ (x)
f ′′ (x)
=5x 4 + 3
20
x 3
=20x 3 +20x 2
=20x 2 (x+1)
Step-by-step explanation:
Example: Finding the inflection points of f(x)=x^5+\dfrac53x^4f(x)=x
5 + 3
5
x 4
f, left parenthesis, x, right parenthesis, equals, x, start superscript, 5, end superscript, plus, start fraction, 5, divided by, 3, end fraction, x, start superscript, 4, end superscript
Step 1: Finding the second derivative
To find the inflection points of fff, we need to use f''f
′′
f, start superscript, prime, prime, end superscript:
\begin{aligned} f'(x)&=5x^4+\dfrac{20}{3}x^3 \\\\ f''(x)&=20x^3+20x^2 \\\\ &=20x^2(x+1) \end{aligned}
f ′ (x)
f ′′ (x)
=5x 4 + 3
20
x 3
=20x 3 +20x 2
=20x 2 (x+1)
Olga was asked to find where f(x)=(x-2)^4f(x)=(x−2)
4
f, left parenthesis, x, right parenthesis, equals, left parenthesis, x, minus, 2, right parenthesis, start superscript, 4, end superscript has inflection points. This is her solution:
Step 1:
\begin{aligned} f'(x)&=4(x-2)^3 \\\\\\ f''(x)&=12(x-2)^2 \end{aligned}
f
′
(x)
f
′′
(x)
=4(x−2)
3
=12(x−2)
2
Step 2: The solution of f''(x)=0f
′′
(x)=0f, start superscript, prime, prime, end superscript, left parenthesis, x, right parenthesis, equals, 0 is x=2x=2x, equals, 2.
Step 3: fff has inflection point at x=2x=2x, equals, 2.