10/7/2021

## 2.4 Connections F, F' And Trigap Calculus

66 ### 2.4 Connections F F' And Trigap Calculus Solutions

25=2!2!2!2!2=32 26=2!2!2!2!2!2=64 c. They are half as large each time. Divided by 2, or multiplied by ½ is also acceptable. 20=2!1 2 =1,!2'1=1!1 2 =1 2,2'2=1 2!1 2 =1 2,!2'3=1 4!1 2 =1 8,!2'4=1 8!1 2 =1 16 2'n=1 2n'1!1 21 =1 2n'1+1 =1 2n 1-9. 2!4'22=2!2!!!Check:!1 16 '4=1 4 b. 2!1'2!2=2!3Check:!1 2 '1 4 =1 8 c. So f prime of a is equal to 0. And if f prime of a is equal to 0 and if we're concave upwards in the interval around a, so if the second derivative is greater than 0, then it's pretty clear, you see here, that we are dealing with a minimum point at a. Name: Block: Date: AP Calculus 4.3c Even More Connecting f ′ and f ′ ′ with the Graph of f Example 1 More on moving from the graph of f ′ to draw a possible graph of f a) The graph of f ′ is shown below (Note: This is the graph of the derivative of f) a) Over what intervals is the graph of f increasing? Whenever lim x → a f (x) is in the domain of F and the composition F (f (x)) makes sense. The reason for which (1.2) happens is in fact a more general (at least formally). ### 2.4 Connections F F' And Trigap Calculus Pdf

According to the product rule, if f(x)=h(x)g(x), then f'(x)=h(x)g'(x)+g(x)h'(x). If we set h(x)=3\$x^{2}\$+6, and g(x)=2x-\$frac{1}{4}\$, then using the product rule, f'(x)=(3\$x^{2}\$+6)\$frac{d}{dx}\$[2x-\$frac{1}{4}\$]+(2x-\$frac{1}{4}\$)\$frac{d}{dx}\$[3\$x^{2}\$+6] \$frac{d}{dx}\$[2x-\$frac{1}{4}\$]=2, using the power rule \$frac{d}{dx}\$[3\$x^{2}\$+6]=6x, using the power rule Therefore, f'(x)=(3\$x^{2}\$+6)(2)+(6x)(2x-\$frac{1}{4}\$) = 6\$x^{2}\$+12+12\$x^{2}\$-\$frac{3x}{2}\$ =18\$x^{2}\$-\$frac{3x}{2}\$+12