Find the function f given that the slope of the tangent line at any point (x, f(x)) is f '(x) and that the graph of f passes through the given point.&nbsp;f'(x)=4(2x−9)3&nbsp;5,11/2

f′(x)=4(2x−9)3∫f′(x)=∫4(2x−9)3dx=4∫(2x−9)3dxu=2x−9⇒du=2dx⇒dx=du2f(u)=4∫u3du2=42∫u3du=2u44=12u4+C⇒f(u)=12u4+Cf(x)=12(2x−9)4+Cx=5,f(x)=112f(x)=12(2x−9)4+C112=12(2[5]−9)4+C112=12(10−9)4+C112=12(1)4+C112=12+CC=112−12=102=5

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Amiya Arellano

Answered question

2022-02-14

Find the function f given that the slope of the tangent line at any point (x, f(x)) is f '(x) and that the graph of f passes through the given point. $f' (x) = 4 (2 x - 9)^{3} 5, 11 / 2$

Answer & Explanation

Veschettijib

Beginner2022-02-15Added 12 answers

f^{'} (x) = 4 {(2 x - 9)}^{3}

\int f^{'} (x) = \int 4 {(2 x - 9)}^{3} d x = 4 \int {(2 x - 9)}^{3} d x

u = 2 x - 9 \Rightarrow d u = 2 d x \Rightarrow d x = \frac{d u}{2}

f (u) = 4 \int u^{3} \frac{d u}{2} = \frac{4}{2} \int u^{3} d u = \frac{2 u^{4}}{4} = \frac{1}{2} u^{4} + C

\Rightarrow f (u) = \frac{1}{2} u^{4} + C

f (x) = \frac{1}{2} {(2 x - 9)}^{4} + C

x = 5, f (x) = \frac{11}{2}

f (x) = \frac{1}{2} {(2 x - 9)}^{4} + C

\frac{11}{2} = \frac{1}{2} {(2 [5] - 9)}^{4} + C

\frac{11}{2} = \frac{1}{2} {(10 - 9)}^{4} + C

\frac{11}{2} = \frac{1}{2} {(1)}^{4} + C

\frac{11}{2} = \frac{1}{2} + C

C = \frac{11}{2} - \frac{1}{2} = \frac{10}{2} = 5

Bleadafat9om

Beginner2022-02-16Added 13 answers

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