Question

find the slope of the line tangent to the graph f(X) =5x^2 + 3x - 4 at x = 4

Answer 1

@jim_thompson5910

Answer 2

\[\Large f(x) = 5x^2 + 3x - 4\]


\[\Large f(x+h) = 5(x+h)^2 + 3(x+h) - 4\]


\[\Large f(x+h) = 5(x^2+2xh+h^2) + 3(x+h) - 4\]


\[\Large f(x+h) = 5x^2+10xh+5h^2 + 3x+3h - 4\]

Answer 3

\[\Large \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\]


\[\Large \lim_{h\to0}\frac{5x^2+10xh+5h^2 + 3x+3h - 4-(5x^2 + 3x - 4)}{h}\]


\[\Large \lim_{h\to0}\frac{5x^2+10xh+5h^2 + 3x+3h - 4-5x^2 - 3x + 4}{h}\]


\[\Large \lim_{h\to0}\frac{10xh+5h^2+3h}{h}\]

Answer 4

\[\Large \lim_{h\to0}\frac{10xh+5h^2+3h}{h}\]

\[\Large \lim_{h\to0}\frac{h(10x+5h+3)}{h}\]

\[\Large \lim_{h\to0}10x+5h+3\]

\[\Large 10x+5(0)+3\]

\[\Large 10x+3\]

Answer 5

find the derivative algebraically

Answer 6

g(x) = 1/x at x = 2

Answer 7

i found the equation of the tangent line. it is y = -1/4 (x-2) + 1/2 but idk how to find the derivate form the tangent line

Answer 8

how did you find the equation of the tangent line?

Answer 9

plugged it into y -y1 = m (x-x1)

Answer 10

its correct right?

Answer 11

how did you know the slope of the tangent line?

Answer 12

i found it by doing the limit

Answer 13

ok let me verify

Answer 14

ok

Answer 15

ok I got

\[\Large g^{\prime}(x)=-\frac{1}{x^2}\]

So

\[\Large g^{\prime}(2)=-\frac{1}{2^2}\]

\[\Large g^{\prime}(2)=-\frac{1}{4}\]

which means that the slope of the tangent line at x = 2 is -1/4

So you are correct there

Answer 16

so what is the derivative

Answer 17

g'(x) is the derivative function

so in general, the derivative of g(x) is

\[\Large g^{\prime}(x)=-\frac{1}{x^2}\]

the derivative at x = 2 is the slope of the tangent line at x = 2 (so you just plug in x = 2 into g'(x))

Answer 18

confused

Answer 19

how do you reduce this

Answer 20

\[(1/(x+h)^{2} - 1/x ^{2}) / h\]

Answer 21

\[3 = (102-100)/(35-y)\]