Question

Let f(x) = 6.4 sin(x) + 5.9 cos(x). What is the maximum and minimum value of thisfunction?

Answer 1

From f(x) = 6.4 sin(x) + 5.9 cos(x)

`\begin{gathered} \text{The maximum value occurs at where }\frac{d\text{ y}}{d\text{ x}}\text{ = 0 } \\ \end{gathered}`

So we will differentiate the function f(x) = 6.4 sin(x) + 5.9 cos(x)

f(x) = 6.4 sin(x) + 5.9 cos(x)

f'(x) = 6.4 cos(x) - 5.9 sin(x)

6.4 cosx - 5.9 sinx = 0

Squaring both sides we have

`\begin{gathered} 6.4\cos x\text{ + 5.9sinx = 0 } \\ \text{recall that cos x = }\sqrt[]{1-sin^2x} \\ 6.4\text{ }\sqrt[]{1-sin^2x}\text{ + 5.9sinx = 0 } \\ 6.4\text{ }\sqrt[]{1-sin^2x}=\text{ -5.9sinx} \\ \text{square both sides } \\ 40.96(1-sin^2x)=34.81sin^2x \\ 40.96-40.96sin^2x\text{ = }34.81sin^2x \\ 40.96=75.77sin^2x \\ \sin ^2x\text{ = 0.541 taking squaroot} \\ \text{sinx = }\sqrt[]{0.541} \\ \sin x\text{ = 0.735} \\ x\text{ = }\sin ^(-1)0.735 \\ x\text{ = 47.35} \end{gathered}`

We have gotten the value for x, now let's find our maximum and minimum values

`\begin{gathered} f(x)\text{ = 6.4sin(47.35) + 5.9cos(47.35)} \\ =\text{ 4.707 + 3.997} \\ =\text{ 8.70}4 \end{gathered}`

Therefore, the maximum value is 8.704 and the minimum value is -8.707

Note that maximum value is just the direct opposite of minimum value