Question

Help please!figure out the "inflection point" of the following functionf(x)=5x³ + 2x² − 3x​

Answer 1

Answer:

x = -2/15.

Step-by-step explanation:

Finding the second derivative:

f'(x) = 15x^2 +4x - 3

f"(x) = 30x + 4

30x + 4 = 0   for a point of inflection

x = -4/30 =  -2/15

So 30x + 4 is negative up to x =  -2/15 and positive thereafter.

So the curve passes from concave downward to concave upward at x = -2/15.

Answer 2

Answer:

$\displaystyle \left( - \frac{2}{15} , \frac{286}{675} \right)$

Step-by-step explanation:

to figure out the infection point

take derivative both sides:

$\displaystyle f'(x) = \frac{d}{dx} {5x}^{3} + 2 {x}^{2} - 3x$

By sum derivation rule we acquire:

$\displaystyle \rm f'(x) = \frac{d}{dx} {5x}^{3} + \frac{d}{dx} 2 {x}^{2} - \frac{d}{dx} 3x$

apply exponent derivation rule which yields:

$\displaystyle f'(x) = {15x}^{2} + 4{x}^{} - 3$

take derivative in both sides once again which yields:

$\rm\displaystyle f''(x) = \frac{d}{dx} {15x}^{2} + \frac{d}{dx} 4{x}^{} - \frac{d}{dx} 3$

remember that, derivative of a constant is always 0 so,

$\rm\displaystyle f''(x) = \frac{d}{dx} {15x}^{2} + \frac{d}{dx} 4{x}^{} - 0$

by exponent derivation rule we acquire:

$\rm\displaystyle f''(x) = {30x} + 4{}^{}$

substitute f''(x) to 0 figure out the x coordinate of the inflection point:

$\rm\displaystyle {30x} + 4{}^{} = 0$

cancel 4 from both sides:

$\rm\displaystyle {30x} = - 4$

divide both sides by 30:

$\rm\displaystyle {x} = - \frac{2}{15}$

now plugin the value of x to the given function to figure out the y coordinate of the inflection point:

$\rm \displaystyle f(x) = {5 \left( - \frac{2}{15} \right) }^{3} + 2 {\left( - \frac{2}{15} \right) }^{2} - 3 \left( - \frac{2}{15} \right)$

By simplifying we acquire:

$\displaystyle f(x) = \frac{286}{675}$

hence,

the coordinates of inflection point are

$\displaystyle \left( - \frac{2}{15} , \frac{286}{675} \right)$