Math

QuestionFind the slope of the tangent line to f(x)=3x2f(x)=3x^{2} at x=3x=3 using the limit definition of the derivative. Evaluate:
1. f(3+h)=f(3+h)=
2. f(3+h)f(3)=f(3+h)-f(3)=
3. f(3+h)f(3)h=\frac{f(3+h)-f(3)}{h}=
4. limh0f(3+h)f(3)h=\lim_{h \rightarrow 0} \frac{f(3+h)-f(3)}{h}=

Then, find f(3)=f^{\prime}(3)=

Studdy Solution

STEP 1

Assumptions1. The function is f(x)=3xf(x)=3x^{} . The point of interest is x=3x=3
3. We are using the limit definition of the derivative

STEP 2

First, we need to find the value of the function at x=+hx=+h. We can do this by substituting +h+h into the function.
f(+h)=(+h)2f(+h)=(+h)^{2}

STEP 3

Expand the expression in the parentheses.
f(3+h)=3((3+h)(3+h))f(3+h)=3((3+h)(3+h))

STEP 4

Calculate the value of f(3+h)f(3+h).
f(3+h)=3(9+6h+h2)f(3+h)=3(9+6h+h^{2})

STEP 5

implify the expression.
f(3+h)=27+18h+3h2f(3+h)=27+18h+3h^{2}

STEP 6

Now, we need to find the value of the function at x=3x=3. We can do this by substituting 33 into the function.
f(3)=3(32)f(3)=3(3^{2})

STEP 7

Calculate the value of f(3)f(3).
f(3)=3(9)=27f(3)=3(9)=27

STEP 8

Now, we need to find the difference between f(3+h)f(3+h) and f(3)f(3).
f(3+h)f(3)=(27+18h+3h2)27f(3+h)-f(3)=(27+18h+3h^{2})-27

STEP 9

implify the expression.
f(3+h)f(3)=18h+3h2f(3+h)-f(3)=18h+3h^{2}

STEP 10

Now, we need to find the ratio of the difference to hh.
f(3+h)f(3)h=18h+3h2h\frac{f(3+h)-f(3)}{h}=\frac{18h+3h^{2}}{h}

STEP 11

implify the expression.
f(3+h)f(3)h=18+3h\frac{f(3+h)-f(3)}{h}=18+3h

STEP 12

Now, we need to find the limit as hh approaches 00.
limh0f(+h)f()h=limh0(18+h)\lim{h \rightarrow0} \frac{f(+h)-f()}{h}=\lim{h \rightarrow0} (18+h)

STEP 13

Calculate the limit.
limh0f(3+h)f(3)h=18\lim{h \rightarrow0} \frac{f(3+h)-f(3)}{h}=18So, f(3)=18f^{\prime}(3)=18.
This means that the slope of the tangent line to the curve f(x)=3x2f(x)=3x^{2} at x=3x=3 is 1818.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord