Answer to Question #119769 in Calculus for Shania kumar

Question

Given the function f(x) = 2xsquared - x-1
(f) List the y - intercepts, if there is any, of the graph of f .
(g) Find all point(s) at which f'(x) =0?
(h) Determine the intervals on which the graph of f (x) is increasing and the intervals on which
it is decreasing.
(i) Use the First Derivative Test to identify the maximum and/or minimum point(s) of f (x) .
(j) Use the Second Derivative Test to determine the nature of concavity of f (x) .

Answer 1

f) Solution

At y-intercept $x=0$

Therefore $y=2(0)^2-(0)-1$

$y=-1$

Answer: The y -intercept is $(0,-1)$

g) Solution

$f(x)=2x^2-x-1$

$f'(x)=4x-1$

At critical point $f'(x)=0$

$4x-1=0$

$4x=1$

$x=0.25$

Answer: At critical point $x=0.25$

h) Solution

From the critical point, we have intervals $(-∞,0.25)$ and $(0.25,∞)$

Let's take $-1$ in interval $(-∞,0.25)$

$f'(-1)=4(-1)-1$

$-4-1=-5$

Therefore $(-∞,0.25)$ is a decreasing interval since $f'(x)$ is negative.

Let's take $1$ in interval $(0.25,∞)$ ;

$f'(1)=4(1)-1$

$=3$

Therefore $(0.25,∞)$ is an increasing interval since $f'(x)$ is positive.

Answer: $f(x)$ is decreasing on interval $(-∞,0.25)$ and increasing on interval $(0.25,∞)$

i) Solution

At maximum or minimum $f'(x)=0$

$f'(x)=4x-1$

$4x-1=0$

$x=0.25$

Substitute $-1,0.25$ and $1$ in $f(x)$ to determine whether $x=0.25$ is maximum or minimum

$f(-1)=2(-1)^2-(-1)-1$

$f(-1)=2$

$f(0.25)=2(0.25)^2-(0.25)-1$

$f(0.25)=-1.125$

$f(1)=2(1)^2-(1)-1$

$f(1)=0$

Answer: From above it is clear that $f(x)$ is minimum at $x=0.25$

j) Solution

$f'(x)=4x-1$

$f''(x)=4$

Answer: Since the second derivative is positive the function is concave up.