Question

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of ( y = f(x) ).

( f(x)=4 x^{6}-9 x^{5} )

the function ( f ) has a local maximum at ( x = )
(type an exact answer, using radicals as needed. use a comma to separate answers as needed.)

the function ( f ) has no local maximum.

select the correct choice below and, if necessary, fill in the answer box to complete your choice.

the function ( f ) has a local minimum at ( x=\frac{15}{8} )
(type an exact answer, using radicals as needed. type an integer or a fraction. use a comma to separate answers as needed.)

the function ( f ) has no local minimum.

select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a. the graph of ( f ) is concave upward on the subinterval(s)
(type your answer in interval notation. type an exact answer, using radicals as needed. type an integer or a fraction. use a comma to separate answers as needed.)

b. the function ( f ) is never concave upward

Explanation:

Step1: Find the first derivative

Use the power rule \((x^n)^\prime=nx^{n - 1}\).
For \(y = f(x)=4x^{6}-9x^{5}\), the first derivative \(f^\prime(x)=24x^{5}-45x^{4}=3x^{4}(8x - 15)\).
Set \(f^\prime(x)=0\), then \(3x^{4}(8x - 15)=0\).
Solving \(3x^{4}(8x - 15)=0\) gives \(x = 0\) or \(x=\frac{15}{8}\).
Use the first - derivative test.

• When \(x\lt\frac{15}{8}(x

eq0)\), let \(x = 1\), then \(f^\prime(1)=24\times1^{5}-45\times1^{4}=24 - 45=-21\lt0\).

• When \(x\gt\frac{15}{8}\), let \(x = 2\), then \(f^\prime(2)=24\times2^{5}-45\times2^{4}=24\times32-45\times16=768 - 720 = 48\gt0\).

Since the function changes from decreasing (\(f^\prime(x)\lt0\)) to increasing (\(f^\prime(x)\gt0\)) at \(x=\frac{15}{8}\), \(x = \frac{15}{8}\) is a local minimum.
Since \(f^\prime(x)\) does not change sign at \(x = 0\) (the sign of \(f^\prime(x)\) is negative on both sides of \(x = 0\) for \(x
eq0\)), there is no local maximum.

Step2: Find the second derivative

Differentiate \(f^\prime(x)=24x^{5}-45x^{4}\) using the power rule.
\(f^{\prime\prime}(x)=120x^{4}-180x^{3}=60x^{3}(2x - 3)\).
Set \(f^{\prime\prime}(x)=0\), then \(60x^{3}(2x - 3)=0\).
Solving \(60x^{3}(2x - 3)=0\) gives \(x = 0\) or \(x=\frac{3}{2}\).

• When \(x\lt0\), let \(x=-1\), then \(f^{\prime\prime}(-1)=120\times(- 1)^{4}-180\times(-1)^{3}=120 + 180=300\gt0\).
• When \(0\lt x\lt\frac{3}{2}\), let \(x = 1\), then \(f^{\prime\prime}(1)=120\times1^{4}-180\times1^{3}=120 - 180=-60\lt0\).
• When \(x\gt\frac{3}{2}\), let \(x = 2\), then \(f^{\prime\prime}(2)=120\times2^{4}-180\times2^{3}=120\times16-180\times8=1920 - 1440 = 480\gt0\).

Answer:

The function \(f\) has no local maximum.
The function \(f\) has a local minimum at \(x=\frac{15}{8}\).
The graph of \(f\) is concave upward on the sub - intervals \((-\infty,0)\cup(\frac{3}{2},\infty)\)