For the function 𝑓(𝑥) = 𝑎𝑥^4+ 𝑏𝑥^3− 4𝑥^2+ 2𝑐𝑥 + 14, determine the values of a, b, and c so that 𝑓(−2) = 2, 𝑓′(−2) = 16 and 𝑓′′(−2) = −8

Answer 1

If f(x)=ax^4+bx^3-4x^2+2cx+14, then

f(-2)=a(-2)^4+b(-2)^3-4(-2)^2+2c(-2)+14

=16a-8b-16-4c+14=16a-8b-4c-2.

or

f(-2)=16a-8b-4c-2.

Since we are given the condition that f(-2)=2, we have the equation:

16a-8b-4c-2=2

Simplifying as much as we can by adding two to both sides:

16a-8b-4c=4

Dividing both sides by 4:

Equation 1: 4a-2b-c=1.

Next condition, f'(-2)=16.

If f(x)=ax^4+bx^3-4x^2+2cx+14, then f'(x)=4ax^3+3bx^2-8x+2c by taking the derivative using the power rule (x^n)'=nx^(n-1).

Then f'(-2)=4a(-2)^3+3b(-2)^2-8(-2)+2c=4a(-8)+3b(4)+16+2c=-32a+12b+2c+16.

Since we are given the condition f'(-2)=16, we have the equation -32a+12b+2c+16=16

Simplifying this by subtracting 16 from both sides, we get:

-32a+12b+2c=0.

Dividing both sides by -2:

Equation 2: 16a-6b-c=0.

Next condition, f''(-2)=-8.

Since f'(x)=4ax^3+3bx^2-8x+2c from the previous part, taking the derivative again using the power rule gives:

f''(x)=12ax^2+6bx-8.

Then f''(-2)=12a(-2)^2+6b(-2)-8=48a-12b-8.

The condition f''(-2)=-8 yields the equation:

48a-12b-8=-8.

Adding 8 to both sides:

48a-12b=0.

Dividing both sides by 12:

4a-b=0

Add b to both sides:

Equation 3: 4a=b

Summarizing our results:

Equation 1: 4a-2b-c=1

Equation 2: 16a-6b-c=0

Equation 3: 4a=b

Then substituting Equation 3 (b=4a) into Equation 1 gives:

4a-4b-c=1

4a-2(4a)-c=1

4a-8a-c=1

-4a-c=1

Equation 4: 4a+c=-1

Substituting Equation 3 (b=4a) into Equation 2 gives:

16a-6b-c=0

16a-6(4a)-c=0

16a-24a-c=0

-8a-c=0

Equation 5: c=-8a

Summarizing our results again:

Equation 4: 4a+c=-1

Equation 5: c=-8a

Plugging Equation 5 (c=-8a) into Equation 4 gives:

4a+c=-1

4a+(-8a)=-1

-4a=-1

a=1/4.

Now since we have a=1/4, plugging that into Equation 5 gives c=-8a=-8(1/4)=-2 or c=-2.

Finally, since we have a=1/4, we can plug that into Equation 3: b=4a=4(1/4)=1 or b=1.

Summarizine everything:

a=1/4, b=1, c=-2,

so the f(x) that satisfies those conditions is

f(x)=ax^4+bx^3-4x^2+2cx+14

f(x)=(1/4)x^4+(1)x^3-4x^2+2(-2)x+14

f(x)=(1/4)x^4+(1)x^3-4x^2-4x+14

You can check your answer by finding f(-2), f'(-2), and f''(-2) and check that they equal 2, 16, and -8 respectively, as I will do now:

f(x)=1/4 x^4+x^3-4x^2-4x+14

plugging in x=-2

f(-2)=1/4(-2)^4+1(-2)^3-4(-2)^2-4(-2)+14=1/4(16)-8-16+8+14=4-8-16+8+14=2. f(-2)=2. Check.

f(x)=1/4 x^4+x^3-4x^2-4x+14

taking the derivative

f'(x)= x^3+3x^2-8x-4

plugging in x=-2

f'(-2)= (-2)^3+3(-2)^2-8(-2)-4=(-8)+3(4)+16-4=-8+12+16-4=16. f'(-2)=16. Check.

f'(x)=x^3+3x^2-8x-4

take the derivative again

f''(x)=3x^2+6x-8

plug in x=-2

f''(-2)=3(-2)^2+6(-2)-8=3(4)-12-8=12-12-8=-8. f''(-2)=-8. Check.