Question

Gr 12 calculus find the critical points on the given interval and identify the nature of them

Answer 1

To find the critical points, take the derivative and set it equal to zero. Use the second derivative test to determine the nature of the critical points.

Step-by-step explanation:

To find the critical points on the given interval, we need to first find the derivative of the function and set it equal to zero. The critical points will be the x-values that make the derivative zero. To identify the nature of these critical points, we can use the second derivative test. If the second derivative is positive at a critical point, then the point is a local minimum. If the second derivative is negative, then the point is a local maximum. If the second derivative is zero or undefined, the test is inconclusive.

Answer 2

we have the function

`f\left(x\right)=2^(x)\cos x`

Find out the critical points

so

Find out the first derivative

`f^(\prime)(x)=ln2*2^xcosx-2^xsinx`

Equate the first derivative to zero

`\begin{gathered} ln2*2^xcosx-2^xsinx=0 \\ ln2=(2^xsinx)/(2^xcosx) \\ \\ ln2=tanx \end{gathered}`

the value of the tangent is positive

that means

the angle x lies on the I quadrant or III quadrant

but remember that the interval is [0, pi]

therefore

The angle x lies on the quadrant

`\begin{gathered} tanx=ln2 \\ x=0.2\pi\text{ radians} \end{gathered}`

The critical point is x=0.2pi radians

Find out the second derivative

`f^(\prime)^(\prime)(x)=ln^22*2^xcosx-ln2*2^(x+1)*sinx-2^xcosx`

Evaluate the second derivative at x=0.2pi radians

The value of the second derivative is negative

so

The concavity is down

that means

The critical point is a local maximum in the given interval